A cricketer throws a ball vertically upwards so that the ball leaves his hands at a speed of 25 m/s. Calculate the maximum height reached by the ball, the time taken to reach max. height, and the speed of the ball when it is at 50% max. height.

We can see that this is a suvat question, since it involves a projectile. The first step for any suvat question is to list the known suvat variables for the point we're looking at (in this case, the point of maximum height):

s = h (we do not yet know the maximum height, h)

u = 25 m/s (initial velocity is given in the question)

v = 0 m/s (velocity at maximum height is zero, because here the ball must change direction from upwards to downwards)

a = -9.8 m/s² (acceleration is always -9.8 m/s² for an object free in the air - it is negative because the gravity causing it always acts downwards towards the Earth's centre)

t = ? (we do not yet know the time taken to reach maximum height)

We can then decide which suvat equation to use from which variables are known. In this case, we do not know t, so the best equation to use is v² = u² + 2as, which does not have t in it. Substitute in the numbers:

0 = 25² + 2(-9.8)(h) = 625 - 19.6h

h = 32 m

Now that we know s, u, v and a, we could use any of the equations to find t. We will choose v = u + at for simplicity:

0 = 25 + (-9.8)(t) = 25 - 9.8t

t = 2.6 s

We then repeat the process for the final part of the question:

s = 0.5 x 32 = 16 ; u = 25; v = ?; a = -9.8; t = ?

Again, we do not know t at 50% of the maximum height, so we will use v² = u² + 2as:

v² = 25² + 2(-9.8)(16) = 311

v = 18 m/s

This gives us the answers h = 32 m, t = 2.6 s, v = 18 m/s.