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Question

# A ball is thrown vertically up from a tower of height 60 m with a speed of 20 m/s. The vâˆ’t curve of the ball is [Take upward positive and g=10 m/s2]

A
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Solution

## The correct option is B Given, height of the building h=60 m, initial speed of the ball u=20 m/s The velocity of the ball is given as v=u+at⇒ v=20−10t So, at t=2 sec, the ball will be at maximum height and velocity will be zero. Now, we will find the time of flight of the ball using the equation s=ut+12at2 and T=2ug So, time taken to cover building height is given by ⇒ −60=−20t−12×10×t2 On solving the equation we get t=2 sec Also, time taken by ball to come at the same level from where it was thrown is given by T=2ug=2×2010=4 sec So, the ball will take 2+4=6 sec to reach the ground Hence, the striking velocity of the ball will be v=20−10×6=−40 m/s Thus, the situation can be shown as The v−t graph would be

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