Answer
5.79 m/s
Work Step by Step
Let's apply the equation $S=ut+\frac{1}{2}at^{2}$ in the vertical direction to find the flight time.
$\uparrow S=ut+\frac{1}{2}at^{2}$ ; Let's plug known values into this equation.
$2.1\space m=15sin50^{\circ}m/s \space t+\frac{1}{2}(-9.8\space m/s^{2})t^{2}$
$4.9\space m/s^{2}t^{2}-11.5\space m/s \space t+2.1\space m=0$
This is a quadratic equation, so let's apply the quadratic formula to find the t.
$t=\frac{-(-11.5\space m/s)\space\pm \sqrt {(-11.5\space m/s)^{2}-4(4.9\space m/s^{2})(2.1\space m)}}{9.8\space m/s^{2}}$
$t=0.2\space s\space or \space 2.145s$
The first root corresponds to the time required for the ball to reach a vertical displacement of 2.1 m as it travels upward, and the second root corresponds to the time required for the ball to have a vertical displacement of 2.1 m as the ball travels downward. So the desired flight time t is 2.145 s.
Let's apply the equation $S=ut$ in the horizontal direction to find the horizontal distance.
$\rightarrow S=ut$ ; Let's plug known values into this equation.
$S=15cos50^{\circ}m/s\times2.145\space s=20.68\space m$
So the opponent must move 20.68 m - 10 m = 10.68 m in 2.145 s - 0.3 s =1.845 s
The minimum average speed of the opponent = $\frac{10.68\space m}{1.845\space s}=5.79\space m/s$
