Question

During a baseball game, a batter hits a high pop-up. If the ball remains in the air for 6.22 s, how high does it rise? The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.

Answer 1

Answer:

47.4 m

Explanation:

When an object is thrown upward, it rises up, it reaches its maximum height, and then it goes down. The time at which it reaches its maximum height is half the total time of flight.

In this case, the time of flight is 6.22 s, so the time the ball takes to reach the maximum height is

$t=\frac{6.22}{2}=3.11 s$

Now we consider only the downward motion of the ball: it is a free fall motion, so we can find the vertical displacement by using the suvat equation

$s=ut+\frac{1}{2}gt^2$

where

s is the vertical displacement

u = 0 is the initial velocity

t = 3.11 s is the time

$g=9.8 m/s^2$ is the acceleration of gravity (taking downward as positive direction)

Solving the  formula, we find

$s=\frac{1}{2}(9.8)(3.11)^2=47.4 m$