You are to throw a ball with a speed of 16.5 m/s at a target that is height h = 4.90 m above the level at which you release the ball. You want the ball's velocity to be horizontal at the instant it reaches the target.
- What is the speed of the ball just as it reaches the target?
Voy = Vo sin (θ) = 36.44°.
V^2 = Vo^2 - 2a(∆y)
The Attempt at a Solution
I thought Vox = Vx (velocity at the top since Vy = 0) = Vo cos (θ) = 5.0599m/s but that was not right so I tried to solve for t using Vy = Voy +at . . 0 = 16sin(36.44) + (-9.8)t and t came out to be 1.6s so i plugged T=1.6s into x = x0 + Vox(t) . . .13.27 = 0 + Vox(1.6) and got 8.29m/s and that too, is wrong. . . What am missing from me getting the correct answer?