Question

While scrambling directly toward the sideline at a speed v_Q = 24 ft/sec, the football quarterback Q throws a pass toward the stationary receiver R. At what angle a should the quarterback release the ball? The speed of the ball relative to the quarterback is 51 ft/sec. Treat the problem as two-dimensional.

Answer 1

the v vector is the velocity of the ball relative to the quarterback → so v(BQ) = 51 ft/s

the vQ vector on the picture is the velocity of the quarterback relative to the ground ... so mark it as v(QG)

then v(QG) = 24 ft/s

The vector that runs towards the receiver is the velocity of the ball relative to the ground v(BG)

so v(BG) = V(BQ) + v(QG)


Now on the picture draw the v(QG) vector from the head of the V(BQ) vector parallel to the vQ vector in the picture down to the line segment QR ... mark the point of intersection I

then v(BG) is the vector that runs from Q to the point of intersection I


Using alternate angles the angle between vQ and QR = the angle between v(QG) and IQ ... call it x

using the marked distances on the picture:

tan x = 18/7.5

so x = arctan (40/4.1) =84.14° ... mark it in the vector triangle

they give that v(QG) = 24 ft/s ... mark it on the vector triangle

and they give that v(BQ) = 51 ft/s ... mark it on the vector triangle


Now we have to find angle α ... the angle between v(BQ) and v(BG) in the vector triangle ... so use the law of sines:

sin α / 24 = sin 84.14° / 51
α = arcsin [24 sin 84.14° / 51] = 27.91° = 28° (to 2 sig figs)