A tennis ball located 2 m height is hit by a player with his racket. The ball shoots out horizontally with a velocity of 30 m/s. Answer the following questions:

a) How long does the ball takes to hit the ground?
b) What is the angle of the velocity vector with the x-axis at the time the ball reaches the ground?
c) If before being hit, the ball is 5 meters away from the net, how high does the ball pass above the net?


Question a)

The ball will reach the ground when the y-coordinate of its position is 0 (y=0). According to the equations of horizontal launch:

y=H-12·g·t20=2 m-0.5·9.8 ms2·t2t=0.63 s

Question b)

To calculate the angle that the velocity vector forms with the x-axis, we will use the following expression:


To solve it, we will calculate vx and vy:

vx=v0=30 msvy=-g·t vy=-9.8 ms2·0.63s=-6.17 ms

Once we have the velocity data, we can get the angle:

tanα=-6.17 ms30 mstanα=-0.205 α=-11.62°


Question c)

To calculate the height at which the ball will pass above the net, first, we must know when it crosses above the net. To do so, knowing that the ball is 5 m away from the net and moves with a u.r.m. at 30 m/s:

x=v0·t 5 m=30 ms·t t=0.17 s

At that time the ball will be just over the net, so we simply calculate its position and we would have solved the problem:

y=H-12·g·t22-0.5·9.8 ms2·(0.17 s)2y=1.86 m

Formulas worksheet

These are the main formulas that you must know to solve this exercise. If you are not clear about their meaning, we recommend you to check the theory in the corresponding sections. Furthermore, you will find in them, under the Formulas tab, the codes that will allow you to integrate these equations in external programs like Word or Mathematica.