A football quarterback throws a pass at an angle of 31.5 degrees. He releases th

Question

A football quarterback throws a pass at an angle of 31.5 degrees. He releases the pass 3.50 m behind the line of scrimmage. His receiver left the line of scrimmage 2.8 s earlier, going straight down-field at a constant speed of 7.50 m/s. With what speed must the quarterback throw the ball so that the pass lands gently in the receiver's hands without the receiver breaking stride? Assume that the ball is released at the same height it is caught and that the receiver is straight downfield from the quarterback at the time of release. (Ignore any effects due to air resistance.)

I am unable to find the velocity that the quarterback throws the football.

Explanation / Answer

So it's basically projectile motion. You have to split the x and y axis. Y(t) = -4.9t^2+v0*t*sin theta+y0 x(t) = v0*t+x0 v0 = initial velocity t = time a = -9.8 theta = angle of release 0 = -4.9t^2+v0*t*(sin31.5) x(t)_ball = v0*cos theta*time = v0*(cos 31.5)*t x(t)_runner = 7.5(t+2.8)+3.5 ball and runner need to travel the same horizontal distance 7.5(t+2.8)+3.5 = v0*(cos 31.5)*t v0 = (7.5*(t+2.8)+3.5) / ((cos31.5)*t)) Plug this isn't the y equation to solve for time. 0 = -4.9t^2+((7.5*(t+2.8)+3.5) / ((cos31.5)*t)))*t*(sin31.5) t = 2.28115 Obviously solved with a machine Go back to y(t) 0 = -4.9t^2+v0*t*(sin31.5) 0 = -4.9*(2.28115)^2+v0*(2.28115)*(sin 31.5) v0 = 21.3927 ^^^^Good variation on projectile motion. If it's wrong, give me some insight. I think the basic approach is right.

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