Advice for the Quarterback question?
A quarterback is preparing to throw the football to a receiver who is running away from him at a constant velocity ( V_r ). The receiver is currently a distance ( D ) away from the quarterback. To ensure the receiver can catch the ball without interference from opposing players, the quarterback estimates that the ball must be thrown at an angle ( A ) to the horizontal and that the receiver must catch the ball after a time interval ( T_c ) from the moment it is thrown.
For this problem, assume that the ball is thrown and caught at the same height above the level playing field. The vertical position of the ball at the moment it is thrown or caught is ( y = 0 ), and the horizontal position of the quarterback is ( x = 0 ). Use ( g ) to represent the magnitude of the acceleration due to gravity, and refer to the provided inertial coordinate system when solving the problem.
a) Determine ( V_{oy} ), the vertical component of the ball's velocity at the moment the quarterback releases it. Express your answer for ( V_{oy} ) in terms of ( T_c ) and ( g ).
b) Calculate ( V_{ox} ), the initial horizontal component of the ball's velocity. Express your answer for ( V_{ox} ) in terms of ( D ), ( T_c ), and ( V_r ).
c) Find the speed ( V_0 ) with which the quarterback must throw the ball. Provide your answer in terms of ( V_r ), ( g ), ( T_c ), and ( D ).
d) Assuming that the quarterback throws the ball with speed ( V_0 ), determine the angle ( A ) above the horizontal at which he should throw it. Your solution should include an inverse trigonometric function (to be entered as ( \text{asin} ), ( \text{acos} ), or ( \text{atan} )). Present your answer in terms of the previously defined quantities ( V_{ox} ), ( V_{oy} ), and ( V_0 ).
1 Answers
(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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