Serving at a speed of 170 km/h, a tennis player hits the ball at a height of 2.5 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? Will the ball land in the service box, whose out line is 6.40 m from the net?

6.11°

yes  

Explanation:

v = 170 km/h = 47.2 m/s

vx = v cos θ

vy = v sin θ

In horizontal direction, there is no acceleration.

vx t = 11.9

In vertical direction:

2.5 - 0.91 = vy t + 0.5 gt²

⇒1.59 = 11.9 tan θ + 4.9×0.25²× sec²θ            (∵g = 9.8 m/s²)

⇒1.59 = 11.9 tan θ + 4.9×0.25²× (1+tan²θ)

Solving for θ will give the result 6.11°

Displacement in x-direction

t = 0.375 s

vx t = 11.9

47.2 cos (6.11°) 0.375 s = 17.6 -11.9 = 5.7 m

The outline is 6.40 m from the net. Thus, the ball will fall in the service box.

positive reinforcement? ?