Question

The point from where a ball is projected is taken as the origin of co-ordinate system. The components of its displacement along x and y are given by x = 6t and $y=8t-5t^2$. What is the velocity and angle of projection?

Answer 1

Let the ball be projected with a velocity $u$ at an angle $\theta$ from the horizontal. At time t, the x and y coordinates will be given by
$x=ucos\theta t$
$y=usin\theta t-\frac{1}{2}g{t}^2=usin\theta t-5t^2$

Comparing with this, we get
$ucos\theta =6$
and $usin\theta =8$

Solving, we get $\theta =ta{n}^{-1}\frac{4}{3}$ and $v=10m/s$.