Question

At a baseball game, the batter hit a fly ball at time t = 0 s. The outfielder caught the ball at t = 5.8 s. When was the ball at maximum elevation? Assume that the baseball was at the same height above the ground when it was hit and caught, and that air resistance is negligible.

Answer 1

We have the following equation for height:
 h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
 Where,
 a: acceleration
 vo: initial speed
 h0: initial height.
 The value of the acceleration is:
 a = -g = -9.8 m / s ^ 2
 For t = 0 we have:
 h (0) = (1/2) * (a) * 0 ^ 2 + vo * 0 + h0
 h (0) = h0
 h0 = 0 (reference system equal to zero when the ball is hit).
 For t = 5.8 we have:
 h (5.8) = (1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0
 (1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0 = 0
 vo = (1/2) * (9.8) * (5.8)
 vo = 28.42
 Substituting values we have:
 h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
 h (t) = (1/2) * (- 9.8) * t ^ 2 + 28.42 * t + 0
 Rewriting:
 h (t) = -4.9 * t ^ 2 + 28.42 * t
 The maximum height occurs when:
 h '(t) = -9.8 * t + 28.42
 -9.8 * t + 28.42 = 0
 t = 28.42 / 9.8
 t = 2.9 seconds.
 Answer:
 The ball was at maximum elevation when:
 t = 2.9 seconds.

Answer 2

Answer:

2.9 s

Explanation:

Vertical Displacement is zero because the initial and final height above the ground is same.

Total time of flight is = 5.8 s

The time taken by ball to reach the maximum elevation is same as the time to fall back the same height. The time at which the ball was at maximum elevation is half the time of flight.

Thus, the time at which the ball was at maximum elevation = $\frac{5.8s}{2}= 2.9s$