A motor cyclist is trying to jump across a path as shown by driving horizontally off a cliff A at a speed of `5ms^(-1)`. Ignore air resistance and take `g=10ms^(-2)`. The speed with which he touches the peak B is:

Options

A. `20 m s^-1`
B. `12 m s^-1`
C. `25 ms^-1`
D. `15 m s^-1`

Answer

D

Explanation

(d) Speed in horizontal direction remains constant during whole journey becomes there is no acceleration in this direction. So, `v_h =m 5 ms^-1`
In vertical direction , loss in gravitation potential energy = gain in ke, i.e.,
`mgh = (1)/(2) mv_V^2`
`V_V^2 = 2gh = 2 xx 10 xx (70 - 60) = 200`
Hence, the speed with which he touches the cliff `B` is :
`v = sqrt(v_h^2 + v_V^2) =sqrt(25 +200) = sqrt(225) = 15 ms^-1`.