An outfielder throws a baseball to his catcher in an attempt to throw out a runn

Question

An outfielder throws a baseball to his catcher in an attempt to throw out a runner at home plate. The ball bounces once before reaching the catcher. Assume the angle at which the bounced ball leaves the ground is the same as the angle at which the outfielder threw it as shown in the figure, but that the ball's speed after the bounce is one-half of what it was before the bounce a)Assuming the ball is always thrown with the same initial speed, at what angle ? should the fielder throw the ball to make it go the same distance D with one bounce (blue path) as a ball thrown upward at 40.0

Explanation / Answer

follow this for the no-bounce the time to apogee is 1/2 the total flight time of the ball vy(t)=v0*sin(35)-g*t when vy(t)=0 the ball is at apogee 0=v0*sin(35)-g*t v0*sin(35)/g=t plug 2x that t into x(t)=v0*cos(35)*t or x(D)=v0^2*sin(35)*cos(35)/g apply the same principles to the bounce call the displacement at the bounce x(b) x(b)=v0^2*sin(?)*cos(?)/g since it bounces at the same angle the distance from the bounce to D is x(b2)=v0^2*sin(?)*cos(?)/(4*g) x(b)+x(b2)=x(D) x(D)=v0^2*sin(?)*cos(?)/g+v0^2*sin(?)*… or 1.25*v0^2*sin(?)*cos(?)/g set equal 1.25*v0^2*sin(?)*cos(?)/g=v0^2*sin(35)… simplify 1.25*sin(?)*cos(?)=sin(35)*cos(35) using 2*sin(?)*cos(?)=sin(2*?) sin(2*?)=sin(70)/1.25 solve for ? 24.37 degrees Ratio of times for the no-bounce 2*v0*sin(35)/g=t for the bounce 2*v0*sin(24.37)/g+2*v0*sin(24.37)/(2*g… simplify 1.5*v0*sin(24.37)/g=t ratio is 1.5*sin(24.37)/sin(35) ratio=1.08 The one bounce takes longer

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