A tennis player aims to serve the ball horizontally when their racquet is 2.5 m

Question

A tennis player aims to serve the ball horizontally when their racquet is 2.5 m above the ground. The distance from the player to the net is 15.0 m and the net is 0.9 m high
(a) What is the minimum speed with which the ball must leave the racquet if it is to just clear the net?
(b) How far beyond the net does the ball land?
(c) If the distance from the net to the service line is 7.0 m is the ball “in” or “out’?
(d) What is the total time for which the ball is in the air from the time that it leaves the racquet until it hits the ground?

Explanation / Answer

a) The ball may fall a maximum of 2.5m-0.9m = 1.6m y = vt + 0.5at^2 = 0 + 0.5at^2; t = sqrt(2y/a) = sqrt(2*1.6/9.8) = 0.57+s So the ball has 0.57s to travel 15.0m: v = 15.0m/0.57+s = 26.25m/s b) The total time the ball travels before it hits the ground (falls 2.5m) is: t = sqrt(2y/a) = sqrt(2*2.5/9.8) = 0.71+s So the total distance travelled horizontally is: x = vt = 26.25*0.71+s = 18.75m c) The ball lands 18.75 - 15.0 = 3.75m from the net. The ball is in. d) We calculated his in part b already. From raquet to ground, the total time is about 0.71 seconds.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.