Hello, KingV15!

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:

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

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

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

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

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

*
* * \
* *\
* *
--*--------→*-------------------*--
B 13.5 C 14.5 A
: - - - - - - 28 - - - - - - :

The kicker at $\displaystyle A$ punts the ball to point $\displaystyle C$, 14.5 m away.

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

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