Two tall buildings are 40m apart. With what speed a ball must be thrown horizontally from a window 145m above the ground in one building, so that it will enter a window 22.5m above the ground in the other ?
(Take g=10m/s2)
A
5ms−1
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B
8ms−1
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C
10ms−1
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D
16ms−1
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Solution
The correct option is B8ms−1
Let the hprizontal velocity of ball be v
Vertical distance between the two windows =h=145−22.5=122.5m
Time taken by the ball to reach from one window to another =t=√2hg
Therefore, horizontal distance covered x=vxt ⇒40=v√2×122.510 ⇒40≈v×5 ⇒v≈8ms−1