Question

A baseball is popped straight up into the air and has an initial velocity of 30.7 m/s. Determine the peak height to which the ball rises.

Answer 1

The peak height to which the ball rises is 48.086 m.

Given:

u= initial velocity= 30.7 m/s

v= final velocity= 0 m/s

To find:

The peak height to which the ball rises (s).

Solution:

Firstly, we have to use the formula v=u+at to find the time taken.

Here, a= g= 9.8 m/s² as the ball is thrown into the air gravitational acceleration will act on it.

Now, put the values of v, u and g in the formula v=u+gt.

0=30.7+(-9.8)t            (g is negative as the ball is thrown upwards)

9.8t=30.7

t= $\frac{30.7}{9.8}$

t= 3.13 sec

Now, to find the peak height to which the ball rises (s) we have to use the formula:

s= ut + 1/2 gt²

Put the values of u, t and g in the above formula.

s= 30.7 ×3.13 - (1/2) × 9.8 × (3.13)²

s= 96.09- 48.004

s= 48.086 m

Therefore, the peak height to which the ball rises is 48.086 m.

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