Question

a body is projected from height of 60 M with a velocity 10 M per second at angle 30° to horizontal the time of flight of the body is

Answer 1

Breaking velocity into components:
$v_{x}=10cos30=5 \sqrt{3}$
$v_{y}=10sin30=5$

By the end of the journey, the final displacement in y of the object will be 60m downwards
Therefore,
 -60=$v_{y}t-5t^{2}$
$-60=5t-5t^{2}$
$-12=t-t^{2}$
$t^{2}-t-12=0$

By solving this equation, we will get 2 values, 4 and -3, since time cannot be negative we will the discard the negative value and consider the positive one instead.

Hence, Time= 4 seconds