Advice for the Quarterback question?

A quarterback is set up to throw the football to a receiver who is running with a constant velocity Vr directly away from the quarterback and is now a distance D away from the quarterback. The quarterback figures that the ball must be thrown at an angle A to the horizontal and he estimates that the receiver must catch the ball a time interval Tc after it is thrown to avoid having opposition players prevent the receiver from making the catch. In the following you may assume that the ball is thrown and caught at the same height above the level playing field. Assume that the y coordinate of the ball at the instant it is thrown or caught is y=0 and that the horizontal position of the quaterback is x=0.

Use g for the magnitude of the acceleration due to gravity, and use the pictured inertial coordinate system when solving the problem.

a) Find Voy, the vertical component of the velocity of the ball when the quarterback releases it.

Express your answer for Voy in terms of Tc, and g.

b) Find Vox, the initial horizontal component of velocity of the ball.

Express your answer for Vox in terms of D, Tc, and Vr.

c)Find the speed Vo with which the quarterback must throw the ball.

answer in temrs of vr g Tc and D

d) Assuming that the quarterback throws the ball with speed vo, find the angle A above the horizontal at which he should throw it.

Your solution should contain an inverse trig function (entered as asin, acos, or atan). Give your answer in terms of already known quantities Vox, Voy, Vo.

1 Answer

  • (a)♣ the vertical component of ball’s velocity is Vy=Voy –g*t;

    at vertex t=Tc/2 and Vy=0= Voy –g*(Tc/2), hence Voy=0.5g*Tc;

    (b)♠ the R-guy catches and his position is x=D+Vr*Tc;

    horizontal position of the ball when caught is x=Vox*Tc,

    where Vox is constant horizontal component of ball’s velocity;

    therefore D+Vr*Tc = Vox*Tc, hence Vox= D/Tc+Vr;

    (c)♣ (Vo)^2 =(Vox)^2 +(Voy)^2, where Vo is initial and final speed of the ball,

    Vo =√((D/Tc +Vr)^2 +0.25(g*Tc)^2);

    (d)♠ Vox=Vo*cosA, Voy=Vo*sinA;

    Voy/Vox = (Vo*sinA) / (Vo*cosA) = tanA, hence

    launch angle is A=atan(Voy/Vox) =atan((0.5g*Tc) /(D/Tc+Vr));

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