A ball is thrown upward with an initial velocity of 100ms−1. Find the height of the ball after 15s. Take g=10ms−2
A
375m
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B
425m
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C
395m
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D
275m
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Solution
The correct option is A375m Here, u=100ms−1,a=−g=−10ms−2 At highest point, v=0 As v=u+gt∴0=100−10×t ∴ Time taken to reach highest point, t=10010=10s The ball will return to the ground at t=20s Corresponding velocity - time graph of the ball is shown in Height attained after 15s=Area of ΔAOB+Area of ΔBCD =500+12(15−10)×(−50)=500−125=375m