Question

A tennis player hits a ball 2.0m above the ground. The ball leaves his racquet with a speed of 20m/s at an angle of 5.0 degrees above the horizontal. The horizontal distance to the net is 7.0m and the net is 1.0m high. Does the ball clear the net? If so, by how much? If not, by how much does it miss.

Answer 1

The ball clears the net.

Height of clearance = 1.008 m

Step-by-step explanation:

Consider the horizontal motion of ball

We have equation of motion s = ut + 0.5at²

Initial velocity, u = 20 cos 5 = 19.92 m/s

Acceleration, a = 0 m/s²

Displacement,s = 7 m

Substituting

s = ut + 0.5at²

7 = 19.92 x t + 0.5 x 0 x t²

t = 0.35 seconds

After 0.35 seconds ball reaches at net place.

Consider the vertical motion of ball

We have equation of motion s = ut + 0.5at²

Initial velocity, u = 20 sin 5 = 1.74 m/s

Acceleration, a = -9.81 m/s²

Time, t = 0.35 s

Substituting

s = ut + 0.5at²

s = 1.74 x 0.35 + 0.5 x -9.81 x 0.35²

s = 0.008 m

Displacement with respect to ground = 2+0.008 = 2.008 m

Height of net = 1 m

So the ball clears the net.

Height of clearance = 2.008 - 1 = 1.008 m