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Question

A ball is thrown up with a speed of 15 m/s. How high will it go before it begins to fall? (g = 9.8 m/s2)

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Solution

The initial velocity of the ball, u = 15 m/s

Acceleration = -g

Final velocity, v = 0 m/s

Let the maximum height attained by the ball before it begins to fall back = H

Using the third equation of motion, v2=u2-2gH

0=(15)2-2×9.8×H19.6H=(15)2H=(15)219.6=22519.6=11.48 mH = 11.48 m

Hence, the maximum height attained by the ball is 11.48 m.

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