i need help on solving a problem

Question

During a baseball game, a batter hits a high pop-up. If the ball remains in the air for 4.72 s, how high above the point where it hits the bat does it rise? Assume when it hits the ground it hits at exactly the level of the bat. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.

Answer 1

s = ut + 0.5at² (u = initial vertical velocity, a = -9.8 m/s/s)

So when s = 0, t = 0, or u + 0.5 a t = 0 so u = -0.5 a t = -0.5 x - 9.8 x 4.72 = 23.128 m/s

As the flight path is a symmetric parabola it will be at the highest point at time 4.72 / 2 = 2.36 s

From s = ut + 0.5 at^2, , you can find the required height substituting u = 23.128, a = -9.8, t = 2.36

= 27.29 m

Mike