Abaseball 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. (hint: the
time to rise to the peak is one-half the total hang-time.)

We have that for the 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. (Hint: the  time to rise to the peak is one-half the total hang-time.)" it can be said that Hence, the height to which the ball rises before it reaches its peak is

S=47.85m

From the question we are told

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. (Hint: the  time to rise to the peak is one-half the total hang-time.)

Generally the Newtons equation for the distance   is mathematically given as

Therefore

S=47.85m

Hence, the height to which the ball rises before it reaches its peak is

S=47.85m

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47.9 m

Explanation:

Given:

y₀ = 0 m

t = 6.25/2 = 3.125 s

v = 0 m/s

a = -9.8 m/s²

Find: y

y = y₀ + vt − ½ at²

y = 0 + (0)(3.125) − ½ (-9.8)(3.125)²

y = 47.9 m

e) none of the above answers are correct

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