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Question

# A stone is thrown from the ground just crosses a wall of height 5 m in 2 s. If the wall is at the horizontal distance 25 m, from the point of projection of the stone, find at what distance the stone falls behind the wall.(Take g=10 ms−2)

A
31.25 m
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B
6.25 m
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C
25 m
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D
32.5 m
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Solution

## The correct option is B 6.25 mLet velocity of projectile is u^i+v^j.Given , in time 2 s, Horizontal distance= 20 m, Vertical Distance = 5 m, Using s=v×t, ⇒20=u×2 ⇒u=12.5ms−1Using s=ut−12gt2 ⇒5=v×2−12g×22 ⇒v=12.5ms−1, At maximum height , Vertical velocity= 0 , Time taken by stone to reach at maximum height is t.⇒0=v−gt ⇒t=1.25s∴Time of Flight=2×t=2.5s, Range R=u×Time of Flight ⇒R=12.5×2.5=31.25m, So the distance between the wall and final point of stone = 31.25-25= 6.25 m.

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