Physics 130
West Chester University

Syllabus || Schedule || Problems || Labs

Lesson 5
2D Kinematics

Note: if you see funny symbols, your browser may not be standards compliant, or you may not have the proper fonts. To get these fonts, click here.

Lesson 6

Example Problems
(These are from Cutnell & Johnson, pp 76 and 77).
In the first problem, we are confronted with the scenario of a tennis game in which you lob a ball over your opponent's head. Your goal is to find how fast he needs to move backwards in order to return the ball. You are given the following pieces of information:

• v_0 = 15 m/s (the speed of your shot)
• \theta = 50.0^o (the angle of your shot above the horizontal
• \Delta x_1 = 10.0 m (the distance between you and your opponentas you hit the ball)
• \Delta t_1 = 0.30 s (the reaction time of your opponent)
• \Delta y = 2.10 m (the distance above the ground at which your opponent makes the return shot) We also define \Delta x_2 as the distance he needs to travel to hit the ball, and \Delta x as the distance travelled by the ball whilst in the air.

To solve this, we need to go in reverse. To get your opponent's speed, you need to know how far he'll go (\Delta x_2), and the time it takes him to do it. To know how far he'll go, you need to know how far the ball goes (\Delta x_2 = \Delta x - \Delta x_1). To find the range (\Delta x) you need to know the time it takes for the ball to fall down. This is almost the same time it takes for the opponent to get back to the ball and hit it (The difference being that it must be adjusted for reaction time). To get the fall time we need to know the initial y velocity and then we can use the quadratic equation.

v_{oy} = v_o*sin \theta

v_{oy} = 15 m/s \times sin(50.0^o) = 11.49 m/s

Now rearranging the standard kinematic equation:
\Delta y = v_{oy} \Delta t + {1 \over 2} a_y \Delta t ^2

so that it is in quadratic form:
{1 \over 2} a_y \Delta t ^2 + v_{oy} \Delta t - \Delta y = 0

We can use the quadratic formula:
\Delta t = {{-v_{oy} \pm \sqrt{v_{oy}^2 + 2 a_y \Delta y}} \over a_y}

\Delta t = {{-11.49 m/s \pm \sqrt{(11.49 m/s)^2 - 2(9.8 m/s^2)(2.10m)}} \over {-9.8 m/s^2}}

The answers are: \Delta t = 0.20 s or 2.14 s Now we can find the range:
\Delta x = v_x \Delta t = v_o cos \theta \Delta t

\Delta x = (15 m/s) cos (50.0^o)(2.14 s)

\Delta x = 20.6 m

Why did we choose the time of 2.14 s instead of 0.20 seconds? Plugging the second value into the equation for range should tell you. With such a value, the ball would only go 1.9 m and not even get to your opponent. So, your opponent had to run from his position 10 meters away from you back an additional 10.6 m (\Delta x_2 = 20.6 m - 10.0 m = 10.6 m). He covered this distance in time:
\Delta t' = \Delta t - \Delta t_1 = 2.14 s - 0.30 s = 1.84 s

Thus the velocity of your opponent must be:
v_{x-opp} = {{\Delta x_2} \over {\Delta t'}} = {{10.6 m} \over {1.84 s}} = 5.7 m/s