A football quarterback throws a pass at an angle of 39.9°. He releases the pass

Question

A football quarterback throws a pass at an angle of 39.9°. He releases the pass 3.50 m behind the lineof scrimmage. His receiver left the line of scrimmage 2.5 s earlier, going straight down-field at a constantspeed of 7.50 m/s. With what speed must the quarterback throw theball so that the pass lands gently in the receiver's hands withoutthe receiver breaking stride? Assume that the ball is released atthe same height it is caught and that the receiver is straightdownfield from the quarterback at the time of release. (Ignore anyeffects due to air resistance.)

A:____________m/s

I have worked on this question for 2.5 hr now and can't get theright answer??????? what am i doing wrong

Explanation / Answer

The receiver covers a distance of x which can be calculate x =vxt = 7.5m/s * 2.5s = 18.75m Since the quarterback is releasing the ball 3.5m behind theline of scrimmage, you need to add this onto 18.75m since thisthrow needs to cover this distance in the same amount of time x =22.25m For the quarterback, x = vocos*t and you aresolving for vo vo = x/cos * t = 22.25m/(cos39.9o* 2.5s) = 11.6m/s Hope that helps vo = x/cos * t = 22.25m/(cos39.9o* 2.5s) = 11.6m/s Hope that helps

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