# Catch Runner

• Mar 21st 2010, 03:04 PM
KingV15
Catch Runner
A player kicks a football, giving it an initial velocity of 12 m/s W, at an angle of 40 degrees up from the horizontal. A second player, standing 28 m west of the kicker, immediately starts running to meet the ball. With what velocity must the seond player run in order to catch the ball before it hits the ground?
• Mar 21st 2010, 05:20 PM
skeeter
Quote:

Originally Posted by KingV15
A player kicks a football, giving it an initial velocity of 12 m/s W, at an angle of 40 degrees up from the horizontal. A second player, standing 28 m west of the kicker, immediately starts running to meet the ball. With what velocity must the seond player run in order to catch the ball before it hits the ground?

start by calculating the time the ball is in the air ...

$t = \frac{2v_{0}\sin{\theta}}{g}$
• Mar 21st 2010, 07:33 PM
Soroban
Hello, KingV15!

Quote:

A player kicks a football, giving it an initial velocity of 12 m/s W, at an angle of 40°.
A second player, 28 m west of the kicker, immediately starts running to meet the ball.
With what velocity must the second player run in order to catch the ball before it hits the ground?

We need the Trajectory Formulas:

. . $\begin{array}{ccccc}\text{Horizontal:} & x &=& (12\cos40^o)t \\ \text{Vertical:} & y &=& (12\sin40^o)t - 4.9t^2 \end{array}$

The ball will hit the ground when $y \,=\,0.$

. . $(12\sin40^o)t - 4.9t^2 \:=\:0 \quad\Rightarrow\quad t\left(12\sin40^o - 4.9t\right) \:=\:0$

Hence: . $t = 0\:\text{ or }\:t \:=\:\frac{12\sin40^o}{4.9} \;\approx\;1.574\text{ seconds}$

In 1.574 seconds, the horizontal distance is: . $x \:=\:12\cos40^o(1.574) \:\approx\:14.5\text{ m}$
Code:

                          *                       *      *  \                   *            *\                 *                *     --*--------→*-------------------*--       B  13.5  C        14.5      A       : - - - - - - 28  - - - - - - :
The kicker at $A$ punts the ball to point $C$, 14.5 m away.

The receiver at $B$ must run 13.5 m in 1.574 seconds.

. . His speed must be: . $\frac{14.5}{1.574} \;\approx\;8.58\text{ m/sec}$