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.

Explanation

` v-y = (dy) /(dt) =8-10 t ` and ` v-x = (dx) /(dt) =6 ,`
so velocity at tiem (t) is
` v = sqrt (v_x^2 + v_y^2 ) = sqrt( 6^2 + (8- 10 t)^20`
At ` t=0`, initial velocity
` u= sqrt ( 6^2 +8^2) = 10 ms^(-1)` .