- #1

MisterAvocadoMan

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## Homework Statement

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 below, but that the ball's speed after the bounce is one-half of what it was before the bounce.

(a) Assume the ball is always thrown with the same initial speed and ignore air resistance. At what angle θ should the fielder throw the ball to make it go the same distance D with one bounce as a ball thrown upward at 45.0° with no bounce?

## Homework Equations

I'm not entirely certain on this, but I feel I should be using general kinematic equations

## The Attempt at a Solution

Well to start with I separated the one bounce ball to be ##x_a## and the no bounce ball to be ##x_b##. ##x_a## has ##x_1## and ##x_2## which are the distance to the bounce and then the remaining distance. I set ##x_1 = V_0 cos(\theta)T_1## and ##x_2 = 1/2 V_0 cos(\theta) T_2## then with ##x_1 + x_2 = x_a## I had to figure out what ##x_a## is which I determined via ##x_a = V_0 cos(45°) T ## which then becomes$$x_a = \frac {\sqrt 2V_0 T }2 $$

With that I get $$\frac {\sqrt 2 V_0 T }2 = V_0 cos(\theta)T_1 + \frac 1 2 V_0 cos(\theta) T_2$$

From there I factor ##V_0## and ##cos(\theta)## out of the right side of the equations to get $$\frac {\sqrt 2 V_0 T }2 = V_0 cos(\theta)(T_1 + \frac 1 2 T_2)$$ and this is about where I start to get stumped. I determined ##T## to be $$T = \frac {2 V_0 sin(\theta)} {g}$$ by rearranging the kinematic equation ##y-y_0 = TV_0 - \frac 1 2 g t^2## to solve for ##T## and assuming that ##y## and ##y_0## are both equal to 0 since it is pitched from the origin and lands on the ground at the end. With that plugged into the equation I get $$\frac {\sqrt 2 V_0}2 (\frac {2V_0 sin(\theta)} {g}) = V_0 cos(\theta)(T_1 + \frac 1 2 T_2)$$ Then I do some cancelling so it's not so complicated to get $$ \frac {\sqrt 2 V_0 sin(\theta)} {g} = cos(\theta)(T_1 + \frac 1 2 T_2)$$ And this is where I'm stuck. I don't know how to deal with ##T_1## and ##T_2## and I'm starting to think I may be way off base here. Any guidance would be appreciated, thank you.