Calculating Receiver's Path and Speed to Catch Thrown Ball

[neochris]

A quarter back throws a ball 18 m/s at an angle of 35 degrees above the horizontal. Standing 18 m away is the receiver. How far does the receiver have to go and at what speed must he travel to catch the ball.

Assume that the delta Y is 0.
Assume that the receiver leaves the same time the ball is thrown.
No outside forces.

 

[himanshu121]

What do u infer from \delta y =0

Can u calculate the distance covered by the ball

 

[ShawnD]

Neat question.

First you have to find the time the ball spends in the air based on the Y velocity. Rearrange the distance formula (d = Vi*t + 1/2*a*t^2) or double the rearranged velocity formula (Vf = Vi + at). I personally prefer the velocity one since quadratics make things more difficult.

$$t = 2(\frac{V_f - V_i}{a})$$

In this case, t is the total time in the air, Vf is 0, Vi is the initial velocity of the ball and a is gravity (remember that gravity is negative). Expanded looks like this

$$t = 2(\frac{0 - 18sin(35)}{-9.81})$$

t = 2.1049

Now find the distance traveled with d = vt.

$$d = (18cos(35))(2.1049)$$

d = 31.036

Now as for how far the guy has to run. He's already 18m away but he has to be 31.036m; he has to run 13.036m and he has to be there in 2.1049s. Just divide to get his speed.

Double check my numbers though.

 

[neochris]

Thanks that helps a lot.