Projectile Motion...

Question

Serving at a speed of 217 km/h, a tennis player hits the ball at a height of 3.1 m and an angle θ below the horizontal. The service line is 11.9 m from the net, which is 0.91 m high. What is the angle θ such that the ball just crosses the net?

Answer 1

217 km/h = 217 * 1000 / 3600 m/s = 60.28 m/s

So when the ball is served horizontal velocity is 60.28 cos θ (constant)

Vertical velocity downwards = 60.28 sin θ (this changes because of gravity).

The ball will be above the net at time t = 11.9 / (60.28 cos θ)

s = ut + 0.5 at² We want s to be 3.1 - 0.91 = 2.19 (distance ball has fallen) when the ball is above the net.

u = 60.28 sin θ

t is 11.9 / (60.28 cos θ)

a is 9.8 (acceleration due to gravity)

s = 2.19

So you need to substitute these in to the equation for s, then solve for θ.

Easiest is to set your calculator to degrees and plot the function s - 2.19 , this needs to be zero so using trace you can find the zero from the graph.

Mike