Question

A baseball is popped straight up into the air and has a hang-time of 6.25 s. Determine the height to which the ball rises before it reaches its peak.
A. 48.0 m.
B. 58.0 m.
C. 78.0 m.
D. 88.0 m.

Answer 1

Hint: The motion of the baseball has to be considered in two parts, ascent and descent. We are given the time for ascent + descent, and we must use the fact that time of ascent is equal to time of descent.

Formula used:
Equations of motion to be used here:
1. v = u – gt, and
2. $h = ut - \dfrac{1}{2}gt^2$;
where symbols have their usual meaning.

Complete answer:
First, we are given the total hanging time of the ball = 6.25 s. If we consider only the situation when the ball is ascending to a height h, then the time required to reach the height h would be:
$t = \dfrac{6.25}{2}$s.
The ball is shot up with an initial velocity u and it comes to rest after a height h, so
$0 = u - gt$
can be used from first equation of motion, which gives,
$u = gt$.
We substitute this result in the second equation of motion, we get:
$h = ut - \dfrac{1}{2}gt^2 = gt^2 - \dfrac{1}{2}gt^2 = \dfrac{1}{2}gt^2 $

Keeping the value of $g = 9.8 ms^{-2}$ and t as found before for the case of ascent, we get:
$h = \dfrac{1}{2}(9.8) \left(\dfrac{6.25}{2} \right)^2 = 47.85 m$

Therefore, the height that the ball reaches is 48 meters approximately.

So, the correct answer is “Option A”.

Note:
It is an important concept to remember that the time of ascent is equal to the time of descent. This simple concept made our calculations much easier here otherwise it would have been complicated or the data would have been insufficient for us to solve this question. The only force acting on the ball is the force of gravity and acceleration (or retardation) on the ball is only due to gravity.