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A pebble is thrown vertically upwards fr...

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 acceleration due to gravity is `9.8 m//s^(2)(a)` what is the height of the bridge? (b) with what velocity does that pebble strike the water?

A

4.9 m, 1.47 m/s

B

9.8 m, 14.7 m/s

C

49 m, 1.47 m/s

D

1.47 m, 4.9 m/s

Text Solution

Verified by Experts

Taking the point of projection as origin and
downward direction as positive. By 2nd equation
of motion, i.e. `s = ut + (1/2)"at"^(2)`, We have,
`h = - 4.9 xx 2 (1//2) 9.8 xx 2^(2) = 9.8`
(u is taken to be negative as it is upwards)
From 1st equation of motion i.e. v = u + at,
`v = – 4.9 + 9.8 × 2 = 14.7 m//s`
Hence correct answer is (B)
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