Aim: How can we solve projectile motion (trajectory) problems? Do Now: An object is thrown straight up with a velocity of 10 m/s. How long is it in the air for? v i = 10 m/s a = -9.8 m/s 2 d = 0 m t = ? d y = v iy t + ½a y t 2 t = 2 s

Projectiles Launched at an angle

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UmQ10 UmQ10

Evil Knievel

Water Slide

Projectiles have both x and y components

Horizontal motion: Velocity is constant (the same as horizontally fired objects) Vertical Motion The same as vertically fired objects Horizontal motion – look at the dots Vertical motion – look at the dots Physlet Physics Ch. 3 I.3.4

Initial Velocity Unlike horizontal projectiles, there are both v ix and v iy Initial velocity can be resolved into x and y components vivi v ix = v i cosθ v iy = v i sinθ θ

Can’t remember your trig?! It’s in your reference table!!

Is there an easier way to remember if v ix or v iy is sinθ or cosθ? It’s in your reference table!! In this case, A = v i

Now what? Since you can determine your v ix and v iy, the rest we can figure out!!! Horizontal velocity is still constant Horizontal acceleration is still zero Vertical acceleration is m/s 2 (just like in vertically fired objects) Vertical v f at max height is 0 m/s (just like in vertically fired objects) Vertical displacement is 0 m (the object starts on the ground and ends on the ground) Time to max height = ½ total time

Example You kick a soccer ball into the air at a 30 o angle with a velocity of 30 m/s. Find the time in the air. First step: Find v ix and v iy x y a = -9.8 m/s 2 d = 0 m t = ?

v ix = v i cosθ v ix = (30 m/s)cos30 v ix = 26 m/s v iy = v i sinθ v iy = (30 m/s)sin30 v iy = 15 m/s x y a = -9.8 m/s 2 d = 0 m t = ? v iy = 15 m/s t = ? v ix = 26 m/s d = v i t + ½at 2 0 m = (15 m/s)t + ½(-9.8 m/s 2 )t 2 0 = 15t – 4.9t 2 4.9t 2 = 15t 4.9t = 15 t = 3.1 s

Find the maximum height x y a = -9.8 m/s 2 d = 0 m v iy = 15 m/s t = 3.1 s d max = ? v fmax = 0 m/s t max = 1.55 s v ix = 26 m/s t = 3.1 s Choose a formula: v fmax 2 = v iy 2 + 2ad max d max = v iy t max + ½at max 2 d max = (15 m/s)(1.55 s) + ½(-9.8 m/s 2 )(1.55 s) 2 d max = m

Solve for the range x y a = -9.8 m/s 2 d = 0 m v iy = 15 m/s t = 3.1 s d max = ? v fmax = 0 m/s t max = 1.55 s v ix = 26 m/s t = 3.1 s d = ? Remember: v ix = constant Therefore, v ix = d = 80.6 m

Thinking question What angle gives the greatest range? 45 o

Which angle gives the:greatest range? greatest height? 45° 90°

Another example problem A projectile is fired at an angle of 22 o A projectile is fired at an angle of 22 o with an initial velocity of 120 meters/second. How highwill the projectile travel? How far will the projectile go?

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