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A pebble is thrown vertically upwards from a bridge with an initial velocity of 4.9 m/s. It strikes the water after 2s. If acc. Due to gravity of `9.8 m//s^(2)` (a) what is the height of the bridge ? (b) with what velocity does the pebble strike the water ?

Text Solution

Verified by Experts

Taking the point of projection as origin and downard direction as positive,
(a) By 2 nd equation of motion, i.e., `s = ut + ((1)/(2)) g t^(2)` we have
`h = - 4.9 xx 2 ((1)/(2)) 9.8 xx 2^(2) = 9.8 m`
(`u` is taken to be negative as it is upwards).
(b) 1 st equation of motion, i.e, `v = u + at`
`v = - 4.9 + 9.8 xx 2 = 14.7 m//s`
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