The height `y` and the distance `x` along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by `y = (8t - 5t^2) m` and `x = 6tm`, where `t` is in seconds. The velocity with which the projectile is projected at `t = 0` is.

Options

A. `6 m s^-1`
B. `8 m s^-1`
C. `10 m s^-1`
D. `14 m s^-1`

Answer

C

Explanation

( c) `v_x = (dx)/(dt) = 6 and v_y = (dy)/(dt) = 8 - 10 t`
Putting `t = 0` (Since we have to find initial velocity)
`v_y = 8 - 10 xx 0 = 8`
`v = sqrt(v_x^2 + v_y^2)) = sqrt(6^2 + 8^2) =10 ms^-1`.