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umka21 [38]
3 years ago
7

A ball player catches a ball 3.2 s after throwing it vertically upward. what height did it reach?

Physics
1 answer:
cluponka [151]3 years ago
8 0

At the top of the height, the velocity is zero and acceleration is negative of acceleration due to gravity ( i.e  -9.8 m/s^2).

The time of the ball in air is 3.2 s, so ascending time is 3.2/2=1.6 s.

Therefore from kinematic equation,

v = u + gt

Substituting the values we get,

0= u - 9.8 (1.6)\\\\u=15.7 m/s, Here v = 0 at top.

Now from equation,

h=ut+\frac{1}{2} gt^2,  here h is the height .

So,

h=(15.7 m/s) (1.6s)-\frac{1}{2} 9.8m/s^2(1.6)^2\\\\ h=25.12m-12.54m\\\\h=12.48 m.

Thus, the ball reached at its maximum height of 12.48 m.

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