A football quarterback is moving straight backward at a spee

A football quarterback is moving straight backward at a speed of 2.8 m/s when he throws a pan to a player 19 m straight downfield. The ball is thrown at an angle of 22 degree relative to the ground and is caught at the same height as it is released. What is magnitude of the initial velocity of the ball relative to the quarterback What angle above the horizontal does the initial velocity of the ball relative to the quarterback make? Give your answer in degrees.

Solution

Here, g=9.81 m/s^2

Further, d=d0+vt = 19 + 2.8*t

And, d=vt = v0cos(22)*t

Now, equating the two, we have -
v0cos(22)t= 19 + 2.8*t ----------------------(1)

Again,

h=vyt+1/2gt^2
0= v0sin(22)t-4.9t^2
v0sin(22)t= 4.9t^2 ...(2)

So, dividing (2) by (1), we have -
tan(22)= 4.9t^2/(19 + 2.8*t)
0.40 = 4.9t^2/(19+2.8t)
4.9t^2 - 1.12t - 7.6 = 0

t = [-1.12 + sqrt(1.12^2 +4*4.9*7.6)] / (2*4.9) = 1.136 s (Discarding negative values)

Now, this is the flight duration.

So, v0cos(22)t=19+2.8t

=> v0*cos(22)*1.136 = 19 + 2.8*1.136

=> v0*1.05 = 22.18

=> v0 = 21.12 m/s

Part(a)

Initial velocity of the ball with respect to quarterback = 21.12 - 2.8 = 18.32 m/s

Part (b)

Now, horizontal component of the ball velocity w.r.t. quarterback = 21.12cos22 - 2.8 = 16.78 m/s

vertical component of ball\'s velocity = 21.12sine22 = 7.91 m/s

So, the requisite angle of the ball relative to quarterback = tan inverse(7.91 / 16.78) = 25.24 deg.