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Question

# A ball is thrown vertically upward with a velocity of 10 m/s from the top of a tower of height of 50 m. Find the time taken by the ball to reach a height 10 m from the ground. (Take g=10 m/s2)

A
1 s
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B
3 s
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C
4 s
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D
5 s
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Solution

## The correct option is C 4 sTaking origin O at ground and vertically upward direction as +y axis direction. Initial velocity of ball: uy=10 m/s, ux=0 Acceleration of ball: ay=−g=−10 m/s2, ax=0 With respect to origin O, the displacement of ball in y-direction when it is at a height 10 m from ground is: Sy=S2−S1=−40 m Using kinematic equation in y-direction: Sy=uyt+12ayt2 ⇒−40=10t+12(−10)t2 ⇒5t2−10t−40=0 t2−2t−8=0 ∴(t−4)(t+2)=0 ⇒t=4,−2 Neglecting -ve value of time, we get: t=4 s

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