 A projectile is an object that has been given an initial velocity by some sort of short-lived force, and then moves through the air under the influence of gravity.  Baseballs, stones, or bullets are all examples of projectiles. Topic 2.1 Extended I – Projectile motion  You know that all objects moving through air feel an air resistance (recall sticking your hand out of the window of a moving car).  We call that air resistance drag and will IGNORE it for the time being. The drag force is a complicating factor that we are not yet ready to deal with. FYI: In the analysis that follows, we will NOT consider the drag force. Keep in the back of your mind that what we derive will be precise only in the absence of air. In air, our results will only be a good, first-order APPROXIMATION.

 It is important to note that the acceleration of a projectile is caused by the gravitational attraction of the earth to the object.  Thus, while there IS an acceleration in the y-direc- tion, there NOT an acceleration in the x-direction. Topic 2.1 Extended I – Projectile motion Projectile slowing down in +y dir. Projectile speeding up in -y dir. Projectile moving at constant speed in +x dir. a x = 0 a y = -g

 The kinematic equations are used for projectile motion, because the accelerations are constant.  The acceleration in the x-direction is 0, and the acceleration in the y-direction is -g. Thus Topic 2.1 Extended I – Projectile motion v x = v 0x + a x t x = x 0 + v 0x t + a x t v x 2 = v 0x 2 + 2a x  x x = x 0 + v 0x t v x = v 0x v y = v 0y + a y t y = y 0 + v 0y t + a y t v y 2 = v 0y 2 + 2a y  y y = y 0 + v 0y t - gt v y = v 0y - gt v y 2 = v 0y 2 + 2g  y General kinematic eq'ns in x General kinematic eq'ns in y Projectile eq'ns in x Projectile eq'ns in y Projectile Motion

 Going back to the original baseball, we find the values of v 0x and v 0y from the velocity vector v 0 :  Recall that v is ALWAYS tangent to the path of the projectile. Thus v 0 is tangent to the path at t 0 : Topic 2.1 Extended I – Projectile motion v0v0 00 v 0x = v 0 cos  0 v 0y = v 0 sin  0  We use the velocity triangle to resolve the components of v 0 :  Now our projectile motion equations are complete. v 0x = v 0 cos  0 v 0y = v 0 sin  0 Projectile Motion

Consider a cannonball that is given an initial velocity of 50.0 m/s at an elevation angle of 30.0°. Topic 2.1 Extended I – Projectile motion 30° x y 50 m/s v0v0 (a) What are v 0x and v 0y ? = 50.0 sin 30.0° = 25.0 m/s = 50.0 cos 30.0° = 43.3 m/s v 0y = v 0 sin  0 v 0x = v 0 cos  0

Consider a cannonball that is given an initial velocity of 50.0 m/s at an elevation angle of 30.0°. Topic 2.1 Extended I – Projectile motion (b) What are the tailored equations of motion of the cannonball? x = x 0 + v 0x t v x = v 0x  x = 43.3t  v x = 43.3 y = y 0 + v 0y t - gt v y = v 0y - gt v y 2 = v 0y 2 + 2g  y  y = 25.0t t 2  v y = t  v y 2 = y Tailored Equations

Consider a cannonball that is given an initial velocity of 50.0 m/s at an elevation angle of 30.0°. Topic 2.1 Extended I – Projectile motion (c) What maximum height above the muzzle will the cannonball reach?  The height is determined using the y equations: v y = t 0 = t 9.80t = 25.0 t = 2.55 s  Now substitute: y = 25.0t t 2 y = 25.0(2.55) - 4.9(2.55) 2 y = 31.9 m Method 1: Question: How do we know v y = 0 at the maximum height? Method 2: v y 2 = y 0 2 = y 19.6y = y = 31.9 m FYI: If all you need is y, Method 2 is the best.

Consider a cannonball that is given an initial velocity of 50.0 m/s at an elevation angle of 30.0°. Topic 2.1 Extended I – Projectile motion (d) How long does it take the cannonball to come back down to its original muzzle height?  The time is determined using the y equations: y = 25.0t t 2 0 = 25.0t t 2 Question: How do we know y = 0 at the desired time? 4.90t 2 = 25.0t 4.90t = 25.0 t = 5.10 s Method 1: Method 2: v y 2 = y v y 2 = (0) v y 2 = v y = ±25.0 v y = Question: How do we know v y is NEGATIVE at the desired time? v y = t = t 9.80t = 50.0 t = 5.10 s

Consider a cannonball that is given an initial velocity of 50.0 m/s at an elevation angle of 30.0°. Topic 2.1 Extended I – Projectile motion (e) How far does the cannonball travel before reaching its original muzzle height?  Distance is determined using the x equations:  We know it takes to 5.10 s to reach this distance. x = 43.3t x = 43.3(5.10) x = 221 m

Consider a cannonball that is given an initial velocity of 50.0 m/s at an elevation angle of 30.0°. Topic 2.1 Extended I – Projectile motion (f) In what direction is the cannonball traveling when it reaches its original muzzle height?  Direction is determined using the velocity equations:  We know it takes to 5.10 s to reach this position. v x = 43.3 m/s v y = t v y = (5.10) v y = m/s v x = 43.3 m/s v y = m/s v   = tan -1 vyvxvyvx = tan = 30.0°

Consider a cannonball that is given an initial velocity of 50.0 m/s at an elevation angle of 30.0°. Topic 2.1 Extended I – Projectile motion (g) What is the displacement of the cannonball at t = 2.00 seconds?  Displacement is determined using the position equations: x = 43.3t y = 25.0t t 2 x = 43.3(2.00) x = 86.6 m y = 25.0(2.00) (2.00) 2 y = 30.4 m  x = x - x 0 = 86.6 m  y = y - y 0 = 30.4 m

T HE R ANGE E QUATION Topic 2.1 Extended I – Projectile motion  Sometimes a projectile begins and ends its flight at the same height. If such is the case we can derive a simple formula that tells us the range R of the projectile as a function of initial velocity v 0 and elevation angle (or projection angle)  0. x y v0v0 R θ0θ0  From the projectile equations we get... x = x 0 + v 0x t y = y 0 + v 0y t - gt R = 0 + v 0 (cos θ 0 )t 0 = 0 + v 0 (sin θ 0 )t - gt R = v 0 (cos θ 0 )t gt 2 = 2v 0 (sin θ 0 )t t = 2v 0 (sinθ 0 ) g R = (2 sinθ 0 cosθ 0 ) v02g v02g R = sin 2θ 0 v02g v02g the range equation R = v 0 (cos θ 0 ) 2v 0 (sinθ 0 ) g FYI: Recall the trigonometric identity 2 sin  cos   sin 2  FYI: sin 2  has a maximum value of 1. Therefore R has a maximum value of R max = v 0 2 /g when  0 = 45 °. This equation is only valid if beginning and ending heights are the same.

T HE R ANGE E QUATION Topic 2.1 Extended I – Projectile motion What angle of elevation (projection angle) should a port cannon have in order to hit a pirate ship 800. m off shore. Assume the muzzle velocity of the cannon ball is 150. m/s. the “level” R = 800 m θ0θ0 R = sin2θ 0 v02g v02g 800 = sin 2θ v 0 = 150 m/s FYI: We need to find  0.  From the range equation we get

T HE R ANGE E QUATION Topic 2.1 Extended I – Projectile motion What angle of elevation (projection angle) should a port cannon have in order to hit a pirate ship 800. m off shore. Assume the muzzle velocity of the cannon ball is 150. m/s. 800 = sin 2θ = sin 2θ 0 800(10) sin 2θ 0 = x y 2θ 0 = 20.83° 2θ 0 = 180° ° 2θ 0 = ° θ 0 = 79.59° θ 0 = 10.42° 2θ 0 This angle also has the same sine of One possibility Another possibility opp hyp SOH

T HE R ANGE E QUATION Topic 2.1 Extended I – Projectile motion What angle of elevation (projection angle) should a port cannon have in order to hit a pirate ship 800. m off shore. Assume the muzzle velocity of the cannon ball is 150. m/s. the “level” R = 800 m v0v0 v0v0 θ 0 = 79.59° θ 0 = 10.42°  Which way takes longest?  There are TWO ways to hit the ship.

(a) How long is the stone in the air? Topic 2.1 Extended I – Projectile motion Suppose you throw a stone at a 30.0° angle and a speed of 50.0 m/s from the top of a bridge that is 40.0 m above a river. x y 50 m/s FYI: Since initial and final heights are NOT the same, we CANNOT use the range equation. We will work this problem out from the basic projectile motion equations. 30° x max FYI: Put in a reasonable coordinate system. In this case, if the y-axis begins at the lowest point the ball will ever reach in the problem, you will never have to worry about negative y-values.

(a) How long is the stone in the air? Topic 2.1 Extended I – Projectile motion Suppose you throw a stone at a 30.0° angle and a speed of 50.0 m/s from the top of a bridge that is 40.0 m above a river.  First analyze your velocity triangle: 30° x y 50 m/s v0v0 = 50.0 sin 30.0° = 25.0 m/s = 50.0 cos 30.0° = 43.3 m/s v 0y = v 0 sin  0 v 0x = v 0 cos  0

(a) How long is the stone in the air? Topic 2.1 Extended I – Projectile motion Suppose you throw a stone at a 30.0° angle and a speed of 50.0 m/s from the top of a bridge that is 40.0 m above a river.  The flight time is limited by the y-equations: y = y 0 + v 0y t - gt  0 = t t t t = 0 a b c t = -b ± b 2 - 4ac 2a = = -25 ± (-4.9)(40) 2(-4.9) = -25 ± = = s = 6.38 s Method 1:

(a) How long is the stone in the air? Topic 2.1 Extended I – Projectile motion Suppose you throw a stone at a 30.0° angle and a speed of 50.0 m/s from the top of a bridge that is 40.0 m above a river.  Some people prefer to NOT use the quadratic formula: Method 2: v y 2 = v 0y 2 + 2g  y  v y 2 = (0-40) v y 2 = 1409 v y = FYI: Why MUST you choose the NEGATIVE v y ? v y = v 0y - gt  v y = t = t 9.80t = t = 6.38 s

(b) How far along the river does the stone go? Topic 2.1 Extended I – Projectile motion Suppose you throw a stone at a 30.0° angle and a speed of 50.0 m/s from the top of a bridge that is 40.0 m above a river.  For horizontal distance use the equations in x: x = x 0 + v 0x t  x = 43.3t x = 43.3(6.38) x = 276 m

Topic 2.1 Extended I – Projectile motion Suppose an airplane traveling at 144 km/h at an altitude of 400 m is to make an emergency drop of a package to a group stranded on an island. How far down range must the package be dropped in order to land on the island? x y 0 = 400 m 144 = km h 144 km h 1000 m 1 km 1 h 3600 s = 40 m/s FYI: The original velocity of the projectile is HORIZONTAL, and the same speed as the plane. FYI: Convert the speed of the plane into compatible units (m/s). v 0 = 40 m/s v 0x = 40 m/s v 0y = 0 m/s x 0 = 0 m y 0 = 400 m a x = 0 m/s 2 a y = m/s 2 The initial values

Topic 2.1 Extended I – Projectile motion Suppose an airplane traveling at 144 km/h at an altitude of 400 m is to make an emergency drop of a package to a group stranded on an island. How far down range must the package be dropped in order to land on the island? x y 0 = 400 m v 0 = 40 m/s y = 400 – 5t 2 0 = 400 – 5t 2 5t 2 = 400 t 2 = 80 t = 8.94 s x = 40t x = 40(8.94) = 358 m time limit range FYI: The package must be dropped 358 m down-range of the target.

Topic 2.1 Extended I – Projectile motion