A ball is thrown from a height of $12.5 m$ from the ground level in the horizontal direction. It falls at a horizontal distance of $200 m$ . The initial velocity of the ball (approximately) is

140 m / s

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80 m / s

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126 m / s

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120 m / s

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Solution

The correct option is C $126 m / s$
$s = u t + \frac{1}{2} a t^{2}$
consider this equation in vertical direction.
$\Rightarrow 12.5 = \frac{1}{2} (10) t^{2} \Rightarrow t = \sqrt{2.5} s$
Thus, it requires $\sqrt{2.5} s$ to reach the bottom.
Now, in the horizontal direction,
$v e l o c i t y = \frac{d i s p l a c e m e n t}{t i m e} = \frac{200}{\sqrt{2.5}} = 126.49$

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A ball is thrown from a height of 12.5m from the ground level in the horizontal direction. It falls at a horizontal distance of 200m. The initial velocity of the ball (approximately) is

The correct option is C 126m/ss=ut+12at2consider this equation in vertical direction.⇒12.5=12(10)t2⇒t=√2.5sThus, it requires √2.5s to reach the bottom.Now, in the horizontal direction,velocity=displacementtime=200√2.5=126.49

A ball is thrown from a height of $12.5 m$ from the ground level in the horizontal direction. It falls at a horizontal distance of $200 m$ . The initial velocity of the ball (approximately) is