A body is projected horizontally from the top of a cliff with a velcoityof `19 .6 ms^(-1)`. What time elapses before horizontal and vertical velocities become equal.
` Take `g=9.8 ms^(-2)`.

Explanation

Here, `U_(x) =19.6 m//s ^(-1), u_(y) =0`
`a_(y) =9.8 ms^(-2)`, a_(x) =0`.
Let horizontal and vertical velocities of projectile be same after time (t)
Taking horizontal motion of body for time (t)
`v_(x) =u_(x) + a_(x) t= 19.6 +0 xx t= 19.6 m// s`
Taking vertical downward motion of body for time (t)
`v_(y) - u_(y) + a_(y) t= 0 + 9.8 t= 9.8 (t)`
As per queston , `v_(x) =v_(y) so 19.6 =9.8 (t)`
or `t= 2 s`.