Question

You are the quarterback ready to pass the ball to a receiver already 15.7 m down the field (from you). He is running straight away at a constant 7.6 m/s, and you will release the ball at an angle of 45.3°. With what speed (in m/s) should you throw the ball?

Answer 1

So we know that range in projectile motion is given by:

R = V0x*T = V0*cos A*(2*V0*sin A)/g

R = (V0^2*sin 2A)/g

Given that

A = 45.3 deg and g = 9.81 m/sec^2, So

R = V0^2*(sin 90.6 deg)/9.81

R = 0.102*V0^2

Now this range should be equald to

R = 15.7 + distance traveled by receiver during that time

R = 15.7 + Speed of receiver*Time

R = 15.7 + 7.6*(2*V0*sin A)/g

R = 15.7 + V0*(2*7.6*sin 45.3 deg)/9.81

R = 15.7 + V0*1.10

Since both ranges are equal, So

15.7 + V0*1.10 = 0.102*V0^2

0.102*V0^2 - 1.10*V0 - 15.7 = 0

Solving above quadratic equation

V0 = [1.10 +/- sqrt (1.10^2 + 4*0.102*15.7)]/(2*0.102)

V0 = 18.92 m/sec = initial speed of thrower (you)

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