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.

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.