When baseball outfielders throw the ball, they usually allow it to take one boun

Question

When baseball outfielders throw the ball, they usually allow it to take one bounce, on the theory that the ball arrives at its target sooner that way. Suppose that, after the bounce, the ball rebounds at the same angle that it had when it was released (as in the figure below), but loses half its speed. Assuming the ball is always thrown with the same initial speed, at what angle theta should the ball be thrown in order to go the same distance D with one bounce (blue path) as a ball thrown upward at phi = 33.6 degree with no bounce (green path)? Determine the ratio of the times for the one-bounce and no-bounce throws. t1b/t0b

Explanation / Answer

assume initial speed is v. then in one bounce case: dstance D=v^2*sin(2*33.6)/9.8=0.094*v^2 in two bounce case: let the angle be theta. then total distance in two bounces=v^2*sin(2*theta)/9.8+ (v/2)^2*sin(2*theta)/9.8=(5*v^2/4)*sin(2*theta)/9.8=0.1275*v^2*sin(2*theta) comparing with D, 0.1275*sin(2*theta)=0.094 sin(2*theta)=0.7369 2*theta=47.473 degrees theta=23.736 degrees b)ratio of time=(2*v*sin(theta)/g + 2*(v/2)*sin(theta)/g)/(2*v*sin(33.6)/9.8))=(sin(23.736)+0.5*sin(23.736))/(sin(33.6))=1.1

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