A piece of marble is projected from earth's surface with velocity of `19.6 sqrt(2) ms^(-1)` at `45^(@)`. 2 s later its velocity makes an angle `alpha` with horizontal, where `alpha` is

Options

A. `45^(@)`
B. `30^(@)`
C. `60^(@)`
D. `0^(@)`

Answer

D

Explanation

`u_(x) = 19.6 ms^(-1)` and `u_(y) = 19.6 ms^(-1)`. After 2s, vertical component of velocity `v_(y)` will become zero. So particle is at its highest point.