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Question

# A pebble is thrown vertically upwards from bridge with an initial velocity of 4.9 m/s. It strikes the water after 2s. If acceleration due to gravity is 9.8 m/s2. The height of the bridge & velocity with which the pebble strikes the water will respectively be-

A
4.9 m, 1.47 m/s
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B
9.8 m, 14.7 m/s
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C
49 m, 1.47 m/s
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D
1.47 m, 4.9 m/s
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Solution

## The correct option is B 9.8 m, 14.7 m/sGiven datau=4.9m/sa=−9.8m/st=2sTaking the point of projection as origin and downward direction as positive, By second equation of motion, i.e., s=ut+12at2s=h,aftert=2sh=4.9×2+12(−9.8)(2)2h=9.8−2(9.8)h=−9.8m|h|=9.8mvelocity after 2sBy first equation of motion, i.e.,v=u+atv=4.9+(−9.8)(2)v=−14.7m/s↓

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