Answer to Question #241199 in Physics for Bonbon

Question

Projectile Motion

1. In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is 0.600 m and the acceleration achieved from this position is 1.25 times the
acceleration due to gravity, g. How far can they jump? State your assumptions
(Increased range can be achieved by swinging the arms in the direction of the jump.)

2. Serving at a speed of 170 km/h, a tennis player hits the ball at a height of 2.5 m and an angle o below the horizontal. The service line is 11.9 m from the net, which is 0.91 m high What is the angle @ such that the ball just crosses the net? Will the ball land in the service box, whose outline is 6.40 m from the net?

Answer 1

1. The acceleration upward is

$$a=1.25g=12.25\text{ m/s}^2.$$

The distance the body can accelerate (h=0.6 m) will achieve reaching the upward speed of

$$v=\sqrt{2ah}=3.83\text{ m/s}.$$

The height of jump is

$$H=\frac{v^2}{2g}=0.75\text{ m}.$$

We assume uniform acceleration of the jumper's body here.

2. The speed in m/s:

$$v=170·1000/3600=47.22\text{ m/s}.$$

The ball is

$$y=2.5-0.91=1.59\text{ m}$$

above the net.

The time of the flight to just cross the net:

$$t_u=\frac{v\sin\theta}{g}=7.82\sin\theta,\\\space\\ t_d=\sqrt{\frac{2((v\sin\theta)^2/(2g)+y)}{g}},\\\space\\ T=t_u+t_d=\frac{v\sin\theta}{g}+\sqrt{\frac{2[(v\sin\theta)^2/(2g)+y]}{g}}.$$

On the other hand, it is

$$T=\frac{R}{v\cos\theta}.$$

Equate and solve for the angle:

$$\frac{R}{v\cos\theta}=\frac{v\sin\theta}{g}+\sqrt{\frac{2[(v\sin\theta)^2/(2g)+y]}{g}},\\\space\\ \theta=24°.$$

The ball will not hit the service box.