Projectiles Launched over Level Ground

Projectiles Launched over Level Ground

Suppose a projectile, like a soccer ball, is kicked over level ground with some initial speed v1 at an angle θ to the horizontal

Suppose a projectile, like a soccer ball, is kicked over level ground with some initial speed v1 at an angle θ to the horizontal v1 θ Horizontal, level ground

Suppose a projectile, like a soccer ball, is kicked over level ground with some initial speed v1 at an angle θ to the horizontal What is an expression for the horizontal vector component in terms of v1and θ ? v1 θ Horizontal, level ground

Suppose a projectile, like a soccer ball, is kicked over level ground with some initial speed v1 at an angle θ to the horizontal v1 θ Horizontal, level ground + v1cosƟ

Suppose a projectile, like a soccer ball, is kicked over level ground with some initial speed v1 at an angle θ to the horizontal What is an expression for the vertical vector component in terms of v1and θ ? v1 θ Horizontal, level ground + v1cosƟ

Suppose a projectile, like a soccer ball, is kicked over level ground with some initial speed v1 at an angle θ to the horizontal v1 + v1sinƟ θ Horizontal, level ground + v1cosƟ

We want to derive an expression for time of flight. What motion determines time of flight, vertical or horizontal? v1 + v1sinƟ θ + v1cosƟ Horizontal, level ground

Vertical Motion v1 + v1sinƟ θ + v1cosƟ Horizontal, level ground

Vertical Motion Given: ag = ? Δdy = ? v1y = ? v1 + v1sinƟ θ + v1cosƟ Horizontal, level ground

Vertical Motion Given: ag = -g where g =10 m/s2 Δdy = ? v1y = ? v1 + v1sinƟ θ + v1cosƟ Horizontal, level ground

Vertical Motion Given: ag = -g where g =10 m/s2 Δdy = 0 v1y = ? v1 + v1sinƟ θ + v1cosƟ Horizontal, level ground

Vertical Motion Given: ag = -g where g =10 m/s2 Δdy = 0 v1y = + v1sinƟ v1 + v1sinƟ θ + v1cosƟ Horizontal, level ground

Vertical Motion Given: ag = -g where g =10 m/s2 Unknown: Δdy = 0 v1y = + v1sinƟ v1 + v1sinƟ θ + v1cosƟ Horizontal, level ground

Vertical Motion Given: ag = -g where g =10 m/s2 Unknown: t = ? Δdy = 0 v1y = + v1sinƟ v1 + v1sinƟ θ + v1cosƟ Horizontal, level ground

Vertical Motion Given: ag = -g where g =10 m/s2 Unknown: t = ? Δdy = Formula: v1y = + v1sinƟ v1 + v1sinƟ θ + v1cosƟ Horizontal, level ground

Vertical Motion Given: ag = -g where g =10 m/s2 Unknown: t = ? Δdy = Formula: Δdy = v1y t + at2/2 v1y = + v1sinƟ v1 + v1sinƟ θ + v1cosƟ Horizontal, level ground

Vertical Motion Given: ag = -g where g =10 m/s2 Unknown: t = ? Δdy = Formula: Δdy = v1y t + at2/2 v1y = + v1sinƟ Sub: v1 + v1sinƟ θ + v1cosƟ Horizontal, level ground

Vertical Motion Given: ag = -g where g =10 m/s2 Unknown: t = ? Δdy = Formula: Δdy = v1y t + at2/2 v1y = + v1sinƟ Sub: = v1sinƟ t -gt2/2 v1 + v1sinƟ θ + v1cosƟ Horizontal, level ground

So far, we have = v1sinƟ t -gt2/2 or… v1 + v1sinƟ θ + v1cosƟ Horizontal, level ground

So far, we have = v1sinƟ t -gt2/2 or… v1sinƟ t -gt2/2 = 0 v1 + v1sinƟ θ + v1cosƟ Horizontal, level ground

So far, we have = v1sinƟ t -gt2/2 or… v1sinƟ t -gt2/2 = 0 t (v1sinƟ -gt/2 ) = 0 v1 + v1sinƟ θ + v1cosƟ Horizontal, level ground

So far, we have = v1sinƟ t -gt2/2 or… v1sinƟ t -gt2/2 = 0 t (v1sinƟ -gt/2 ) = 0 v1sinƟ -gt/2 = 0 v1 + v1sinƟ θ + v1cosƟ Horizontal, level ground

So far, we have = v1sinƟ t -gt2/2 or… v1sinƟ t -gt2/2 = 0 t (v1sinƟ -gt/2 ) = 0 v1sinƟ -gt/2 = 0 or… t = ? v1 + v1sinƟ θ + v1cosƟ Horizontal, level ground

So far, we have = v1sinƟ t -gt2/2 or… v1sinƟ t -gt2/2 = 0 t (v1sinƟ -gt/2 ) = 0 v1sinƟ -gt/2 = 0 or… t = 2 v1sinƟ/g v1 + v1sinƟ θ + v1cosƟ Horizontal, level ground

So far, we have = v1sinƟ t -gt2/2 or… v1sinƟ t -gt2/2 = 0 t (v1sinƟ -gt/2 ) = 0 v1sinƟ -gt/2 = 0 or… t = 2 v1sinƟ/g Memorize this! v1 + v1sinƟ θ + v1cosƟ Horizontal, level ground

We want to derive a formula for horizontal displacement of a projectile fired over level ground. What is the only formula we can start with for the horizontal motion of a projectile? v1 + v1sinƟ θ + v1cosƟ Horizontal, level ground Horizontal displacement Δdx ?

Δdx = vxt Why is this the only formula we can use? v1 + v1sinƟ θ + v1cosƟ Horizontal, level ground Horizontal displacement Δdx ?

Δdx = vxt = v1cosƟ t v1 + v1sinƟ θ + v1cosƟ Horizontal, level ground Horizontal displacement Δdx ?

Δdx = vxt = v1cosƟ t = v1cosƟ (2 v1sinƟ/g ) v1 + v1sinƟ θ + v1cosƟ Horizontal, level ground Horizontal displacement Δdx ?

Δdx = vxt Δdx = v12 (2sinƟcosƟ)/g = v1cosƟ t = v1cosƟ (2 v1sinƟ/g ) v1 + v1sinƟ θ + v1cosƟ Horizontal, level ground Horizontal displacement Δdx ?

Δdx = vxt Δdx = v12 (2sinƟcosƟ)/g = v1cosƟ t Δdx= v12 (sin(2Ɵ))/ g = v1cosƟ (2 v1sinƟ/g ) v1 + v1sinƟ θ + v1cosƟ Horizontal, level ground Horizontal displacement Δdx ?

Δdx = vxt Δdx = v12 (2sinƟcosƟ)/g = v1cosƟ t * Δdx= v12 (sin(2Ɵ))/ g = v1cosƟ (2 v1sinƟ/g ) Memorize the above equation v1 + v1sinƟ θ + v1cosƟ Horizontal, level ground Horizontal displacement Δdx ?

Review of Short-Cut Formulas for Projectiles fired over level ground

Time of flight formula?

Time of flight formula? t = 2 v1sinƟ/g

Time of flight formula? t = 2 v1sinƟ/g Horizontal displacement formula?

Time of flight formula? t = 2 v1sinƟ/g Horizontal displacement formula? Δdx= v12 (sin(2Ɵ))/ g

Time of flight formula? t = 2 v1sinƟ/g Horizontal displacement formula? Δdx= v12 (sin(2Ɵ))/ g Memorize these short-cut formulas but beware… Only use them for:

Time of flight formula? t = 2 v1sinƟ/g Horizontal displacement formula? Δdx= v12 (sin(2Ɵ))/ g Memorize these short-cut formulas but beware… Only use them for: Projectiles fired over level ground Δdy= 0

Time of flight formula? t = 2 v1sinƟ/g Horizontal displacement formula? Δdx= v12 (sin(2Ɵ))/ g Memorize these short-cut formulas but beware… Only use them for: Projectiles fired over level ground and Δdy= 0 Do not use these short-cuts if Δdy≠ 0

Time of flight formula? t = 2 v1sinƟ/g Horizontal displacement formula? Δdx= v12 (sin(2Ɵ))/ g Memorize these short-cut formulas but beware… Only use them for: Projectiles fired over level ground and Δdy= 0 Do not use these short-cuts if Δdy≠ 0 Note: Range = ǀ Δdx ǀ

Example #1: A golf ball is fired over level, flat ground with an initial speed of 36.0 m/s and at angle of 36.9° to the ground. Using GUFSA, find: a) time of flight b) the range of the projectile c) maximum height above ground

Example #1: A golf ball is fired over level, flat ground with an initial speed of 36.0 m/s and at angle of 36.9° to the ground. Using GUFSA, find: a) time of flight b) the range of the projectile c) maximum height above ground Given: ? = 36.0 m/s

Example #1: A golf ball is fired over level, flat ground with an initial speed of 36.0 m/s and at angle of 36.9° to the ground. Using GUFSA, find: a) time of flight b) the range of the projectile c) maximum height above ground Given: v1 = 36.0 m/s

Example #1: A golf ball is fired over level, flat ground with an initial speed of 36.0 m/s and at angle of 36.9° to the ground. Using GUFSA, find: a) time of flight b) the range of the projectile c) maximum height above ground Given: v1 = 36.0 m/s ? = 36.9°

Example #1: A golf ball is fired over level, flat ground with an initial speed of 36.0 m/s and at angle of 36.9° to the ground. Using GUFSA, find: a) time of flight b) the range of the projectile c) maximum height above ground Given: v1 = 36.0 m/s Ɵ = 36.9°

Example #1: A golf ball is fired over level, flat ground with an initial speed of 36.0 m/s and at angle of 36.9° to the ground. Using GUFSA, find: a) time of flight b) the range of the projectile c) maximum height above ground Given: v1 = 36.0 m/s Ɵ = 36.9° a) Unknown: = ?

Example #1: A golf ball is fired over level, flat ground with an initial speed of 36.0 m/s and at angle of 36.9° to the ground. Using GUFSA, find: a) time of flight b) the range of the projectile c) maximum height above ground Given: v1 = 36.0 m/s Ɵ = 36.9° a) Unknown: t = ?

Example #1: A golf ball is fired over level, flat ground with an initial speed of 36.0 m/s and at angle of 36.9° to the ground. Using GUFSA, find: a) time of flight b) the range of the projectile c) maximum height above ground Given: v1 = 36.0 m/s Ɵ = 36.9° a) Unknown: t = ? Formula:

Example #1: A golf ball is fired over level, flat ground with an initial speed of 36.0 m/s and at angle of 36.9° to the ground. Using GUFSA, find: a) time of flight b) the range of the projectile c) maximum height above ground Given: v1 = 36.0 m/s Ɵ = 36.9° a) Unknown: t = ? Formula: t = 2 v1sinƟ/g

Example #1: A golf ball is fired over level, flat ground with an initial speed of 36.0 m/s and at angle of 36.9° to the ground. Using GUFSA, find: a) time of flight b) the range of the projectile c) maximum height above ground Given: v1 = 36.0 m/s Ɵ = 36.9° a) Unknown: t = ? Formula: t = 2 v1sinƟ/g Sub:

Example #1: A golf ball is fired over level, flat ground with an initial speed of 36.0 m/s and at angle of 36.9° to the ground. Using GUFSA, find: a) time of flight b) the range of the projectile c) maximum height above ground Given: v1 = 36.0 m/s Ɵ = 36.9° a) Unknown: t = ? Formula: t = 2 v1sinƟ/g Sub: t = 2 (36.0) sin 36.9° /10

Example #1: A golf ball is fired over level, flat ground with an initial speed of 36.0 m/s and at angle of 36.9° to the ground. Using GUFSA, find: a) time of flight b) the range of the projectile c) maximum height above ground Given: v1 = 36.0 m/s Ɵ = 36.9° a) Unknown: t = ? Formula: t = 2 v1sinƟ/g Sub: t = 2 (36.0) sin 36.9° /10 = 4.3 seconds

Example #1: A golf ball is fired over level, flat ground with an initial speed of 36.0 m/s and at angle of 36.9° to the ground. Using GUFSA, find: a) time of flight b) the range of the projectile c) maximum height above ground Given: v1 = 36.0 m/s Ɵ = 36.9° a) Unknown: t = ? Formula: t = 2 v1sinƟ/g Sub: t = 2 (36.0) sin 36.9° /10 = 4.3 seconds b) Unknown: = ?

Example #1: A golf ball is fired over level, flat ground with an initial speed of 36.0 m/s and at angle of 36.9° to the ground. Using GUFSA, find: a) time of flight b) the range of the projectile c) maximum height above ground Given: v1 = 36.0 m/s Ɵ = 36.9° a) Unknown: t = ? Formula: t = 2 v1sinƟ/g Sub: t = 2 (36.0) sin 36.9° /10 = 4.3 seconds b) Unknown: R = ǀΔdxǀ = ?

Example #1: A golf ball is fired over level, flat ground with an initial speed of 36.0 m/s and at angle of 36.9° to the ground. Using GUFSA, find: a) time of flight b) the range of the projectile c) maximum height above ground Given: v1 = 36.0 m/s Ɵ = 36.9° a) Unknown: t = ? Formula: t = 2 v1sinƟ/g Sub: t = 2 (36.0) sin 36.9° /10 = 4.3 seconds b) Unknown: R = ǀΔdxǀ = ? Formula:

Example #1: A golf ball is fired over level, flat ground with an initial speed of 36.0 m/s and at angle of 36.9° to the ground. Using GUFSA, find: a) time of flight b) the range of the projectile c) maximum height above ground Given: v1 = 36.0 m/s Ɵ = 36.9° a) Unknown: t = ? Formula: t = 2 v1sinƟ/g Sub: t = 2 (36.0) sin 36.9° /10 = 4.3 seconds b) Unknown: R = ǀΔdxǀ = ? Formula: Δdx= v12 (sin(2Ɵ))/ g

Example #1: A golf ball is fired over level, flat ground with an initial speed of 36.0 m/s and at angle of 36.9° to the ground. Using GUFSA, find: a) time of flight b) the range of the projectile c) maximum height above ground Given: v1 = 36.0 m/s Ɵ = 36.9° a) Unknown: t = ? Formula: t = 2 v1sinƟ/g Sub: t = 2 (36.0) sin 36.9° /10 = 4.3 seconds b) Unknown: R = ǀΔdxǀ = ? Formula: Δdx= v12 (sin(2Ɵ))/ g Sub:

Example #1: A golf ball is fired over level, flat ground with an initial speed of 36.0 m/s and at angle of 36.9° to the ground. Using GUFSA, find: a) time of flight b) the range of the projectile c) maximum height above ground Given: v1 = 36.0 m/s Ɵ = 36.9° a) Unknown: t = ? Formula: t = 2 v1sinƟ/g Sub: t = 2 (36.0) sin 36.9° /10 = 4.3 seconds b) Unknown: R = ǀΔdxǀ = ? Formula: Δdx= v12 (sin(2Ɵ))/ g Sub: Δdx= (sin(2X36.9°))/ 10

Example #1: A golf ball is fired over level, flat ground with an initial speed of 36.0 m/s and at angle of 36.9° to the ground. Using GUFSA, find: a) time of flight b) the range of the projectile c) maximum height above ground Given: v1 = 36.0 m/s Ɵ = 36.9° a) Unknown: t = ? Formula: t = 2 v1sinƟ/g Sub: t = 2 (36.0) sin 36.9° /10 = 4.3 seconds b) Unknown: R = ǀΔdxǀ = ? Formula: Δdx= v12 (sin(2Ɵ))/ g Sub: Δdx= (sin(2X36.9°))/ 10 = 124 m

Example #1: A golf ball is fired over level, flat ground with an initial speed of 36.0 m/s and at angle of 36.9° to the ground. Using GUFSA, find: a) time of flight b) the range of the projectile c) maximum height above ground Given: v1 = 36.0 m/s c) Unknown: Ɵ = 36.9° a) Unknown: t = ? Formula: t = 2 v1sinƟ/g Sub: t = 2 (36.0) sin 36.9° /10 = 4.3 seconds b) Unknown: R = ǀΔdxǀ = ? Formula: Δdx= v12 (sin(2Ɵ))/ g Sub: Δdx= (sin(2X36.9°))/ 10 = 124 m

Example #1: A golf ball is fired over level, flat ground with an initial speed of 36.0 m/s and at angle of 36.9° to the ground. Using GUFSA, find: a) time of flight b) the range of the projectile c) maximum height above ground Given: v1 = 36.0 m/s c) Unknown: Δdy = ? Ɵ = 36.9° a) Unknown: t = ? Formula: t = 2 v1sinƟ/g Sub: t = 2 (36.0) sin 36.9° /10 = 4.3 seconds b) Unknown: R = ǀΔdxǀ = ? Formula: Δdx= v12 (sin(2Ɵ))/ g Sub: Δdx= (sin(2X36.9°))/ 10 = 124 m

Example #1: A golf ball is fired over level, flat ground with an initial speed of 36.0 m/s and at angle of 36.9° to the ground. Using GUFSA, find: a) time of flight b) the range of the projectile c) maximum height above ground Given: v1 = 36.0 m/s c) Unknown: Δdy = ? Ɵ = 36.9° Why can’t we use short-cut formulas a) Unknown: t = ? over level ground ? Formula: t = 2 v1sinƟ/g Sub: t = 2 (36.0) sin 36.9° /10 = 4.3 seconds b) Unknown: R = ǀΔdxǀ = ? Formula: Δdx= v12 (sin(2Ɵ))/ g Sub: Δdx= (sin(2X36.9°))/ 10 = 124 m

Example #1: A golf ball is fired over level, flat ground with an initial speed of 36.0 m/s and at angle of 36.9° to the ground. Using GUFSA, find: a) time of flight b) the range of the projectile c) maximum height above ground Given: v1 = 36.0 m/s c) Unknown: Δdy = ? Ɵ = 36.9° Δdy ǂ 0 so we must use vertical motion variables! a) Unknown: t = ? Formula: t = 2 v1sinƟ/g Sub: t = 2 (36.0) sin 36.9° /10 = 4.3 seconds b) Unknown: R = ǀΔdxǀ = ? Formula: Δdx= v12 (sin(2Ɵ))/ g Sub: Δdx= (sin(2X36.9°))/ 10 = 124 m

Example #1: A golf ball is fired over level, flat ground with an initial speed of 36.0 m/s and at angle of 36.9° to the ground. Using GUFSA, find: a) time of flight b) the range of the projectile c) maximum height above ground Given: v1 = 36.0 m/s c) Unknown: Δdy = ? Ɵ = 36.9° Δdy ǂ 0 so we must use vertical motion variables! a) Unknown: t = ? Other Vertical motion given: Formula: t = 2 v1sinƟ/g Sub: t = 2 (36.0) sin 36.9° /10 = 4.3 seconds b) Unknown: R = ǀΔdxǀ = ? Formula: Δdx= v12 (sin(2Ɵ))/ g Sub: Δdx= (sin(2X36.9°))/ 10 = 124 m

Example #1: A golf ball is fired over level, flat ground with an initial speed of 36.0 m/s and at angle of 36.9° to the ground. Using GUFSA, find: a) time of flight b) the range of the projectile c) maximum height above ground Given: v1 = 36.0 m/s c) Unknown: Δdy = ? Ɵ = 36.9° Δdy ǂ 0 so we must use vertical motion variables! a) Unknown: t = ? Other Vertical motion given: Formula: t = 2 v1sinƟ/g v1y = ? Sub: t = 2 (36.0) sin 36.9° /10 = 4.3 seconds b) Unknown: R = ǀΔdxǀ = ? Formula: Δdx= v12 (sin(2Ɵ))/ g Sub: Δdx= (sin(2X36.9°))/ 10 = 124 m

Example #1: A golf ball is fired over level, flat ground with an initial speed of 36.0 m/s and at angle of 36.9° to the ground. Using GUFSA, find: a) time of flight b) the range of the projectile c) maximum height above ground Given: v1 = 36.0 m/s c) Unknown: Δdy = ? Ɵ = 36.9° Δdy ǂ 0 so we must use vertical motion variables! a) Unknown: t = ? Other Vertical motion given: Formula: t = 2 v1sinƟ/g v1y = +v1sinƟ Sub: t = 2 (36.0) sin 36.9° /10 = 4.3 seconds b) Unknown: R = ǀΔdxǀ = ? Formula: Δdx= v12 (sin(2Ɵ))/ g Sub: Δdx= (sin(2X36.9°))/ 10 = 124 m

Example #1: A golf ball is fired over level, flat ground with an initial speed of 36.0 m/s and at angle of 36.9° to the ground. Using GUFSA, find: a) time of flight b) the range of the projectile c) maximum height above ground Given: v1 = 36.0 m/s c) Unknown: Δdy = ? Ɵ = 36.9° Δdy ǂ 0 so we must use vertical motion variables! a) Unknown: t = ? Other Vertical motion given: Formula: t = 2 v1sinƟ/g v1y = +v1sinƟ Sub: t = 2 (36.0) sin 36.9° / = 36.0sin 36.9° = 4.3 seconds b) Unknown: R = ǀΔdxǀ = ? Formula: Δdx= v12 (sin(2Ɵ))/ g Sub: Δdx= (sin(2X36.9°))/ 10 = 124 m

Example #1: A golf ball is fired over level, flat ground with an initial speed of 36.0 m/s and at angle of 36.9° to the ground. Using GUFSA, find: a) time of flight b) the range of the projectile c) maximum height above ground Given: v1 = 36.0 m/s c) Unknown: Δdy = ? Ɵ = 36.9° Δdy ǂ 0 so we must use vertical motion variables! a) Unknown: t = ? Other Vertical motion given: Formula: t = 2 v1sinƟ/g v1y = +v1sinƟ Sub: t = 2 (36.0) sin 36.9° / = 36.0sin 36.9° = 4.3 seconds = 21.6 m/s b) Unknown: R = ǀΔdxǀ = ? Formula: Δdx= v12 (sin(2Ɵ))/ g Sub: Δdx= (sin(2X36.9°))/ 10 = 124 m

Example #1: A golf ball is fired over level, flat ground with an initial speed of 36.0 m/s and at angle of 36.9° to the ground. Using GUFSA, find: a) time of flight b) the range of the projectile c) maximum height above ground Given: v1 = 36.0 m/s c) Unknown: Δdy = ? Ɵ = 36.9° Δdy ǂ 0 so we must use vertical motion variables! a) Unknown: t = ? Other Vertical motion given: Formula: t = 2 v1sinƟ/g v1y = +v1sinƟ Sub: t = 2 (36.0) sin 36.9° / = 36.0sin 36.9° = 4.3 seconds = 21.6 m/s b) Unknown: R = ǀΔdxǀ = ? ag = ? Formula: Δdx= v12 (sin(2Ɵ))/ g Sub: Δdx= (sin(2X36.9°))/ 10 = 124 m

Example #1: A golf ball is fired over level, flat ground with an initial speed of 36.0 m/s and at angle of 36.9° to the ground. Using GUFSA, find: a) time of flight b) the range of the projectile c) maximum height above ground Given: v1 = 36.0 m/s c) Unknown: Δdy = ? Ɵ = 36.9° Δdy ǂ 0 so we must use vertical motion variables! a) Unknown: t = ? Other Vertical motion given: Formula: t = 2 v1sinƟ/g v1y = +v1sinƟ Sub: t = 2 (36.0) sin 36.9° / = 36.0sin 36.9° = 4.3 seconds = 21.6 m/s b) Unknown: R = ǀΔdxǀ = ? ag = -10 m/s/s Formula: Δdx= v12 (sin(2Ɵ))/ g Sub: Δdx= (sin(2X36.9°))/ 10 = 124 m

Example #1: A golf ball is fired over level, flat ground with an initial speed of 36.0 m/s and at angle of 36.9° to the ground. Using GUFSA, find: a) time of flight b) the range of the projectile c) maximum height above ground Given: v1 = 36.0 m/s c) Unknown: Δdy = ? Ɵ = 36.9° Δdy ǂ 0 so we must use vertical motion variables! a) Unknown: t = ? Other Vertical motion given: Formula: t = 2 v1sinƟ/g v1y = +v1sinƟ Sub: t = 2 (36.0) sin 36.9° / = 36.0sin 36.9° = 4.3 seconds = 21.6 m/s b) Unknown: R = ǀΔdxǀ = ? ag = -10 m/s/s v2y = 0 m/s Formula: Δdx= v12 (sin(2Ɵ))/ g Sub: Δdx= (sin(2X36.9°))/ 10 = 124 m

Example #1: A golf ball is fired over level, flat ground with an initial speed of 36.0 m/s and at angle of 36.9° to the ground. Using GUFSA, find: a) time of flight b) the range of the projectile c) maximum height above ground Given: v1 = 36.0 m/s c) Unknown: Δdy = ? Ɵ = 36.9° Δdy ǂ 0 so we must use vertical motion variables! a) Unknown: t = ? Other Vertical motion given: Formula: t = 2 v1sinƟ/g v1y = +v1sinƟ Sub: t = 2 (36.0) sin 36.9° / = 36.0sin 36.9° = 4.3 seconds = 21.6 m/s b) Unknown: R = ǀΔdxǀ = ? ag = -10 m/s/s v2y = 0 m/s Formula: Δdx= v12 (sin(2Ɵ))/ g Formula: Sub: Δdx= (sin(2X36.9°))/ 10 = 124 m

Example #1: A golf ball is fired over level, flat ground with an initial speed of 36.0 m/s and at angle of 36.9° to the ground. Using GUFSA, find: a) time of flight b) the range of the projectile c) maximum height above ground Given: v1 = 36.0 m/s c) Unknown: Δdy = ? Ɵ = 36.9° Δdy ǂ 0 so we must use vertical motion variables! a) Unknown: t = ? Other Vertical motion given: Formula: t = 2 v1sinƟ/g v1y = +v1sinƟ Sub: t = 2 (36.0) sin 36.9° / = 36.0sin 36.9° = 4.3 seconds = 21.6 m/s b) Unknown: R = ǀΔdxǀ = ? ag = -10 m/s/s v2y = 0 m/s Formula: Δdx= v12 (sin(2Ɵ))/ g Formula: Sub: Δdx= (sin(2X36.9°))/ Δdy = (v2y 2 - v1y 2)/2 ag = 124 m

Example #1: A golf ball is fired over level, flat ground with an initial speed of 36.0 m/s and at angle of 36.9° to the ground. Using GUFSA, find: a) time of flight b) the range of the projectile c) maximum height above ground Given: v1 = 36.0 m/s c) Unknown: Δdy = ? Ɵ = 36.9° Δdy ǂ 0 so we must use vertical motion variables! a) Unknown: t = ? Other Vertical motion given: Formula: t = 2 v1sinƟ/g v1y = +v1sinƟ Sub: t = 2 (36.0) sin 36.9° / = 36.0sin 36.9° = 4.3 seconds = 21.6 m/s b) Unknown: R = ǀΔdxǀ = ? ag = -10 m/s/s v2y = 0 m/s Formula: Δdx= v12 (sin(2Ɵ))/ g Formula: Sub: Δdx= (sin(2X36.9°))/ Δdy = (v2y 2 - v1y 2)/2 ag = 124 m = (02 – 21.62)/2(-10)

Example #1: A golf ball is fired over level, flat ground with an initial speed of 36.0 m/s and at angle of 36.9° to the ground. Using GUFSA, find: a) time of flight b) the range of the projectile c) maximum height above ground Given: v1 = 36.0 m/s c) Unknown: Δdy = ? Ɵ = 36.9° Δdy ǂ 0 so we must use vertical motion variables! a) Unknown: t = ? Other Vertical motion given: Formula: t = 2 v1sinƟ/g v1y = +v1sinƟ Sub: t = 2 (36.0) sin 36.9° / = 36.0sin 36.9° = 4.3 seconds = 21.6 m/s b) Unknown: R = ǀΔdxǀ = ? ag = -10 m/s/s v2y = 0 m/s Formula: Δdx= v12 (sin(2Ɵ))/ g Formula: Sub: Δdx= (sin(2X36.9°))/ Δdy = (v2y 2 - v1y 2)/2 ag = 124 m = (02 – 21.62)/2(-10) = m

Maximum Range of a Projectile Fired over Level ground

Maximum Range of a Projectile Fired over level ground
What is the formula for range or l∆dxl of a projectile?

Maximum Range of a Projectile fired over level ground
Δdx= v12 (sin(2Ɵ))/ g

Δdx= v12 (sin(2Ɵ))/ g We want to maximize Δdx and thus the range of a projectile. For a given initial launching speed v1 , can you guess what angle Ɵ to the horizontal ground a projectile should be launched to maximize Δdx ?

Δdx= v12 (sin(2Ɵ))/ g We want to know for sure! Let’s look at the expression for Δdx. If we can maximize “sin(2Ɵ)” , we will maximize Δdx.

Δdx= v12 (sin(2Ɵ))/ g What is the maximum value for “sin(2Ɵ)” ?

Δdx= v12 (sin(2Ɵ))/ g The maximum value for “sin(2Ɵ)” is one.

Δdx= v12 (sin(2Ɵ))/ g The maximum value for “sin(2Ɵ)” is one. Remember y = sin(2Ɵ) graph sketch looks like this: +1 -1

Δdx= v12 (sin(2Ɵ))/ g The maximum value for “sin(2Ɵ)” is one. Remember y = sin(2Ɵ) graph sketch looks like this: For what acute angle α is a y=sinα = 1 ? +1 α = ? -1

Δdx= v12 (sin(2Ɵ))/ g The maximum value for “sin(2Ɵ)” is one. Remember y = sin(2Ɵ) graph sketch looks like this: α = 90° +1 α = ? -1

Δdx= v12 (sin(2Ɵ))/ g The maximum value for “sin(2Ɵ)” is one. Remember y = sin(2Ɵ) graph sketch looks like this: α = 90° But what is α in y = sin(2Ɵ) ? +1 α = ? -1

Δdx= v12 (sin(2Ɵ))/ g The maximum value for “sin(2Ɵ)” is one. Remember y = sin(2Ɵ) graph sketch looks like this: α = 90° Ahh.. α=2Ɵ So what is 2Ɵ = ? +1 α = ? -1

Δdx= v12 (sin(2Ɵ))/ g The maximum value for “sin(2Ɵ)” is one. Remember y = sin(2Ɵ) graph sketch looks like this: α = 90° Ahh.. α=2Ɵ So what is 2Ɵ = ? 2Ɵ = 90° +1 α = ? -1

Δdx= v12 (sin(2Ɵ))/ g The maximum value for “sin(2Ɵ)” is one. Remember y = sin(2Ɵ) graph sketch looks like this: α = 90° Ahh.. α=2Ɵ So what is 2Ɵ = ? 2Ɵ = 90° Therefore Ɵ = ? +1 α = ? -1

Δdx= v12 (sin(2Ɵ))/ g The maximum value for “sin(2Ɵ)” is one. Remember y = sin(2Ɵ) graph sketch looks like this: α = 90° Ahh.. α=2Ɵ So what is 2Ɵ = ? 2Ɵ = 90° Therefore Ɵ = 45° to maximize the range!! +1 α = ? -1

Short-Cut Formula for Maximum Range of a Projectile over Level ground

We know the range formula is…

We know the range formula is… ǀΔdxǀ= v12 (sin(2Ɵ))/ g

We know the range formula is… ǀΔdxǀ= v12 (sin(2Ɵ))/ g For maximum range Ɵ = 45° and sin(2Ɵ) = ?

We know the range formula is… ǀΔdxǀ= v12 (sin(2Ɵ))/ g For maximum range Ɵ = 45° and sin(2Ɵ) = 1

We know the range formula is… ǀΔdxǀ= v12 (sin(2Ɵ))/ g For maximum range Ɵ = 45° and sin(2Ɵ) = 1 So what is Rmax = ?

We know the range formula is… ǀΔdxǀ= v12 (sin(2Ɵ))/ g For maximum range Ɵ = 45° and sin(2Ɵ) = 1 So what is Rmax = v12 / g

We know the range formula is… ǀΔdxǀ= v12 (sin(2Ɵ))/ g For maximum range Ɵ = 45° and sin(2Ɵ) = 1 So what is Rmax = v12 / g Memorize!

Homework practice New Textbook:
Read over p41 – p42 Do p42 Q1-Q3 check answers same page Do Q65, Q67 p57 Q72, Q74 p58 harder p59 Q85 check answers p714 Note some of the handout questions can be done using the short-cut formulas Critical Thinking Practice: Try this, we will take up later… For a projectile that lands at the same level from which it starts, state another launch angle above the horizontal that would result in the same range as a projectile launched at an angle of 36° and 16°. Answer: 54°, 74° Old Textbook Read over p46-p49 Try #6 (critical thinking) Q7,Q8 p51 check answers p781 Note some of the handout questions can be done using the short-cut formulas . Harder (take up in class) p67 Q49

Projectiles Launched over Level Ground

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