A ball is thrown upward with an initial velocity of 100ms−1. Draw velocity-time graph for the ball and find from the graph height of the ball after 15s. Take g=10ms−2.
A
125m
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B
375m
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C
500m
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D
625m
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Solution
The correct option is A375m Initial velocity of the ball u=100m/s
Acceleration of the ball a=−g=−10m/s2
Using v=u+at
∴v=100−10t
Thus v−t graph is a straight line intersecting the t axis at t=10 s
Also v∣∣∣t=15=100−10(15)=−50m/s2
Height (displacement) of the ball is equal to the area under the curve during 0<t<15