Horizontal Projectile Launch Constant velocity horizontally Acceleration vertically At any given location, you could solve for total velocity by using pythag. Thm…you first need to know horiz and vert vel.

Horizontal Projectile Launch Suppose v1 = 8 m/s And the ball is 2 m above the floor t = ? b) dx = ? c) Final velocity (at floor) = ? List Givens: Horizontal (x): Vertical (y): ax = ay = vx = vy = dx = dy =

A ball is thrown at 20 m/s at an angle of 35 above horizontal A ball is thrown at 20 m/s at an angle of 35 above horizontal. It lands at the same height as its launch height.

A ball is thrown at 20 m/s at an angle of 35 above horizontal A ball is thrown at 20 m/s at an angle of 35 above horizontal. It lands at the same height as its launch height. t = ? dx = ? a) How long will the ball remain in the air? b) How far will it travel horizontally?

A ball is thrown at 20 m/s at an angle of 35 above horizontal A ball is thrown at 20 m/s at an angle of 35 above horizontal. It lands at the same height as its launch height. t = ? v = ? dy(max) = ? a) How long will it take to get to the max y? b) Describe the velocity at this location c) How high will the ball travel?

Remember…use a vertical equation to solve for time (gravity makes a good timer) 1. Find the y-component of initial velocity v1y = v1 sin 2. Recognize the vertical displacement (dy) dy = 0 3. Use an equation that includes: g, v1, dy, t dy = v1yt + ½ g t2 Solve for t

dy = v1yt + ½ g t2

dx = v1xt v1x = v1 cos 

dy = v1yt + ½ gt2 v2y2 = v1y2 + 2gdy …where t = half the total time. Or use v2y2 = v1y2 + 2gdy such that v2 = 0 More examples

v1 = 25 m/s  = 50 dx = ? v(t) = ? @ t = 1 sec

v1 = 8 m/s  = 30 dy = 1 m dx = ? t = ?

v1 = 45 m/s dy = 10 m  = 40 How far away from the cliff can the thrower stand?