A football is thrown to a moving

Question

football player. The football leaves the quarterback’s
hands 1.6m above the ground with a speed of 14 m/s
at an angle 30o above the horizontal. If the receiver
starts 11 m away from the quarterback along the line
of flight of the ball when it is thrown, what constant
velocity must he have to get to the ball at the instant it
is 1.6m above the ground?

Answer 1

At time t, the ball's height is
h(t) = 1.6 + (14sin30°)t - 4.9t^2
So, the time it takes for the ball to go up and drop back to 1.6 is 1.428 seconds.
The ball's constant horizontal speed is 14cos30° = 12.124 m/s
So, the ball travels 12.124*1.428 = 17.314m
During that time t, the runner's distance is 11+1.428v. So, find v such that
11 + 1.428v = 17.314
v = 4.42 m/s

Answer 2

Since the final height is the same as the initial height call them both 0, not 1.6 to make it easy.
Vi = initial speed up = 14 * sin 30 = 7 m/s upward
when v = 0, we are half way though the problem at the max height.
v = Vi - g t
0 = 7 - 9.8 t
t = .714 m/s at the top
so total time in air = 2 t = 1.43 seconds
how far did it go in that 1.43 seconds?
u = 14 cos 30 = 14 (sqrt 3) / 2 = 7 sqrt 3
so
d = range = 1.43 * 7 sqrt 3
d - 11 = distance that Gronk runs
so
speed times time = d - 11
s * 1.43 = d - 11
s = (d-11)/1.43