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
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
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