Question

A body is projected from height of 60m with a velocity 10 ms-1 at angle 30 degrees to horizontal the time of flight of the body is

Answer 1

When the body hits ground below it would have displaced vertically and horizontally.
The vertical displacement is equal to the initial height of 60 m in the downward direction.
Initial velocity has a vertical upward component of 10 sin 30 = 5 m/s.
Acceleration is due to gravity, so a = 9.8 $ms^{-2}$ in the downward direction.
Lets take upward direction positive and downward negative.

Then initial vertical velocity u = +5 m/s  (upward)
Vertical acceleration a = -9.8 $ms^{-2}$ (downward)
Vertical displacement S = -60 m (downward).
As acceleration is constant we can use $S = ut + \frac{1}{2}at^2$.
Then $-60 = 5t-\frac{1}{2}\times 9.8 t^2$.
Solving this quadratic equation we get t = 4.04 s or -3.02 s.
Discarding the negative value of time, we get t = 4.04 s.