A tennis player hits a ball with an initial speed of 23.6 m/s perfectly horizontally. The ball is 2.37 m above the ground when struck by the racquet. If the net is 12 m from the ball and the net height is 0.9 m, by how much does the ball clear the net?

Refer to the diagram shown below.

Neglect air resistance.

g = 9.8 m/s²

Let t₁ = the time for the ball to travel 12 m horizontally.

That is,

t₁ = (12 m) / (23.6 m/s) = 0.5085 s

The initial vertical launch velocity is zero.

The vertical distance traveled after t₁ is

s = (1/2) * g * (t₁ s) ²

= 0.5*9.8*0.5085²

= 1.267 m

Therefore the height of the ball above ground after t₁ is

2.37 - 1.267 = 1.103 m

The distance between the ball's height and the top of the net is

h = 1.103 - 0.9 = 0.203 m

Answer: 0.203 m