Horizontal Projectile Launch Constant velocity horizontally Acceleration vertically

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 v 1 = 8 m/s And the ball is 2 m above the floor a) t = ? List Givens: Horizontal (x): ax = vx = dx = b) dx = ? b) c) Final velocity =? Vertical (y): ay = vy = dy = (at floor)

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. 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. 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 v 1 y = v 1 sin 2. Recognize the vertical displacement (dy) dy = 0 3. Use an equation that includes: g, v 1, dy, t dy = v 1 yt + ½ g t 2 Solve for t

dy = v 1 yt + ½ g 2 t

dx = v 1 xt v 1 x = v 1 cos

dy = v 1 yt + ½ 2 gt …where t = half the total time. Or use 2 v 2 y = 2 v 1 y + 2 g dy such that v 2 = 0 More examples

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

v 1 = 8 m/s = 30 dy = 1 m t=? dx = ?

v 1 = 45 m/s dy = 10 m = 40 How far away from the cliff can the thrower stand?