Answer to Question #161368 in Mechanics | Relativity for Samir khan

Question

You are on serve in a ping pong match and serve the ball at an angle of 42o above horizontal, 15.0 cm above the end of the table, with a velocity of 6.0m/s. (8 marks total)

a) What is the time of flight? (2 marks)

b) What is the horizontal distance travelled? (1 mark)

c) What is the velocity on impact? (1 mark)

d) If the net is also 15.0 cm high, and 174.0 cm from the edge of the table, will the ball clear the net?

Answer 1

a) We can find the total time of flight of the ping pong from the kinematic equation:

$$y=y_0+v_0tsin\theta-\dfrac{1}{2}gt^2,$$ $$4.9t^2-4.014t-0.15=0.$$

This quadratic eqquation has two roots: $t_1=0.85\ s$ and $t_2=-0.036\ s.$ Since the time can't be negative, the correct answer is $t=0.85\ s.$

b) The  horizontal distance travelled by the ping pong can be found as follows:

$$x=v_0tcos\theta=6.0\ \dfrac{m}{s}\cdot0.85\ s\cdot cos42^{\circ}=3.79\ m.$$

c) The horizontal component of the ping pong's velocity remains unchainged during the flight. Therefore,

$$v_x=v_0cos\theta=6.0\ \dfrac{m}{s}\cdot cos42^{\circ}=4.46\ \dfrac{m}{s}.$$

Let's find the vertical component of the ping pong's velocity:

$$v_y=v_0sin\theta-gt,$$ $$v_y=6.0\ \dfrac{m}{s}\cdot sin42^{\circ}-9.8\ \dfrac{m}{s^2}\cdot0.85\ s=-4.31\ \dfrac{m}{s}.$$

Finally, we can find the velocity on impact from the Pythagorean theorem:

$$v=\sqrt{v_x^2+v_y^2}=\sqrt{(4.46\ \dfrac{m}{s})^2+(-4.31\ \dfrac{m}{s})^2}=6.2\ \dfrac{m}{s}.$$

d) Let's consider the motion of the ping pong in two dimensions:

$$x=v_0tcos\theta,$$ $$y=y_0+v_0tsin\theta-\dfrac{1}{2}gt^2.$$

Let’s find the height of the ping pong as a function of horizontal distance by eliminating the time.

Let’s express time from the first equation and substitute it into the second one:

$$t=\dfrac{x}{v_0cos\theta},$$ $$y=y_0+v_0sin\theta\cdot\dfrac{x}{v_0cos\theta}-\dfrac{1}{2}g(\dfrac{x}{v_0cos\theta})^2,$$ $$y=y_0+xtan\theta-\dfrac{1}{2}g(\dfrac{x}{v_0cos\theta})^2,$$ $$y=0.15\ m+1.74\ m\cdot tan42^{\circ}-\dfrac{1}{2}\cdot9.8\ \dfrac{m}{s^2}\cdot(\dfrac{1.74\ m}{6.0\ \dfrac{m}{s}\cdot cos42^{\circ}})^2,$$ $$y=0.97\ m.$$

Since, the height of the ping pong is greater than the height of the net, the ping pong will clear the net.