Atennis 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

0.203 m

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