A particle is projected up the incline such that its component of velocity along the incline is `10 ms^(-1)`. Time of flight is 4 sec and maximum height above the incline is `5m`. Then velocity of projection will be :

Options

A. `10 ms^(-1)`
B. `10 sqrt2 ms^(-1)`
C. `5 sqrt5 ms^(-1)`
D. None

Answer

C

Explanation

`T = (2u_(y))/(a_(y)) = 4`
`H = (u_(y)^(2))/(2a_(y)) = 5`
`(T)/(H) = (4)/(u_(y)) = (4)/(5)`
`u_(y) = 5`
`u_(x) = 10` (given)
`u = sqrt(10^(2) + 5^(2)) = sqrt125 = 5 sqrt5`