A cricket ball is thrown at a speed of 30ms^-1 in a direction 30 degrees above the horizontal. Calculate:

1. The maximum height.

2. The time taken to return to the same level.

3. The distance from the thrower to where the ball returns to the same level.

Posted this yesterday but still don't understand

1 answer

Read the excellent article at wikipedia:

https://en.wikipedia.org/wiki/Trajectory

You can then understand why

max height = (v sin^2θ)/2g
range = (v^2 sin2θ)/2g

time in air solves
(v sinθ) t - g/2 t^2 = 0