Question

Suppose you now take the ball and using a bat, pop it straight up into the air with a hang-time of 5.00 s (the hang time is how long the baseball is in the air after it leaves the bat). Determine the height to which the ball rises before it reaches its peak.

Answer 1

Answer:

30.66m

Explanation:

Using the equation of motion formula $S = ut + \frac{1}{2}gt^2$ where;

S is the height to which the ball rises

u is the initial velocity of the ball = 0m/s

a is the acceleration due to gravity = 9.81m/s²

t is the time taken by the ball in air = 5.0s

Note that the  time to rise to the peak is one-half the total hang-time = 5.0/2 = 2.5s

Substituting the given parameters into the formula above to get S:

$S = ut + \frac{1}{2}gt^2\\\\S = 0(2.5)+ \frac{1}{2}(9.81)(2.5)^2\\\\S = 0+\frac{1}{2}(9.81)\times 6.25 \\\\S = \frac{61.3125}{2}\\ \\S = 30.65625m\\\\S \approx 30.66m$

This means that the ball rises 30.66m before it reaches its peak.