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

` u_x_0 = ( (dx)/(dt)_(t-0) = d/(dt) (6 t) = 6`
` u_y = ( 9dx)/(dt) = d/(dt) (8 -5t^2) = 8 - 10 t`
` y_y_0 = ((dy)/(dt)_(t-0)= 8 - 10 xx 0= 8 ms^(-1)`
:. ` u= sqrt u_(x_0)^2 + u_y-0^2) = sqrt 6^2+8^2) = 18 ms^(-1)`
If ` theta` is the angle of projection of frjectile, then
` tan theta = u_y_0 /u_x_0 = 8/6 = 4/3 = 1.333 = tan 53^@ 8'`
or ` theta = 53^@ 8'`.