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. `2.0ms^(-1)`
B. `15ms^(-1)`
C. `25ms^(-1)`
D. `20ms^(-1)`

Answer

B

Explanation

Speed in horizontal direction remains constant during whole journey because there is no acceleration in this direction.
So, `v_(h)=6ms^(-1)`
In vertical direction:
Loss of gravitational potential energy =gain in KE
i.e., `mgh=1/2mv_(V)^(2)`
`V_(V)^(2)=2gh=2xx10xx(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)=15ms^(-1)`