# Q. 13.     A ball is thrown vertically upwards with a velocity of $49\; m/s$. Calculate              (i) the maximum height to which it rises,

The initial velocity of the ball u = 49 m s-1

Final velocity at the highest point would be v = 0

The magnitude of the acceleration is equal to the acceleration due to gravity g

Acceleration, a = -g = -9.8 m s-2

Let the maximum height to which it rises be s

Using the third equation of motion we have

$\\v^{2}-u^{2}=2as\\ s=\frac{v^{2}-u^{2}}{2a}\\ s=\frac{0^{2}-49^{2}}{2\times -9.8}\\ s=122.5m$

The ball would rise to a maximum value of 112.5 m.

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