Question

A batter pops a ball straight up (a) If the ball returns to the height from which it was hit 4.5 s later, what was its

Answer 1

Answer:

a)  S = V0 t + 1/2 a t^2     distance ball travels in time t

0 = V0 t - 1\2 * 9.8 * t^2       ball returns to original height

V0 = 4.9 * 4.5 = 22.05 m / s      original speed

b) Since V0 = g t   time to decelerate to zero

t = 22.05 / 9.8 =  2.25 sec       time up equals time down

c) S = V0 t + 1/2 a t^2  

S = 22.05 * 2.25 - 9.8 / 2 * 2.25^2 = 24.8     Max height

Check: Avg speed = 22.05 / 2 = 11.025 m/s

Time to fall from max height = 11.025 * 2.25 = 24.8 ft