# Kinematics: baseball toss with an angle provided

## 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.

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kuruman
The times don't buy you much. I would find the range $R$ in terms of the initial speed for the 45o case and then apply the range equation twice for angle $\theta$ in the equation $x_1+x_2=R$.
The times don't buy you much. I would find the range $R$ in terms of the initial speed for the 45o case and then apply the range equation twice for angle $\theta$ in the equation $x_1+x_2=R$.
Thank you so much for the reply. For some reason the $R$ equation wasn't covered in lecture and I must've overlooked it in the book somehow (we'll chalk that up to me being a dummy haha) I'll make an attempt with that, thank you again.
So following the instructions to use the $R$ equation, I get $R_A = \frac {V_0^2} {g} sin(45*2)$ and since $sin(45*2) = sin(90)$ which equals $1$ I get $\frac {V_0^2} {g}$ which will be the total distance $D$ of the ball toss since this was the range of the ball that doesn't bounce. So for the bounce ball will have two ranges, one being to the point of the first bounce $R_1$ and the other being the remaining distance $R_2$. $R_1$ will be equal to $$R_1 = \frac {V_0^2} {g} sin(2\theta)$$ and $R_2$ will be equal to $$R_2 = \frac {(.5V_0)^2} {g} sin(2\theta)$$ Then with the idea that $R_1 + R_2 = R_A$ since $R_A$ is the full range of the toss, I can write $$\frac {V_0^2} {g} sin(2\theta) + \frac {(.5V_0)^2} {g} sin(2\theta) = \frac {V_0^2} {g}$$ From here I can do some factoring and square up the terms in the parentheses to get $$\frac {V_0^2} {g} sin(2\theta) (1+.25) = \frac {V_0^2} {g}$$ Which will be pretty easy to cancel from to get $$sin(2\theta) = \frac {1} {1.25}$$ which should give me $53.130^{\circ}$ which I'll divide by $2$ to get $26.6^{\circ}$ which seems right for the answer. Thank you again for the guidance, it really helped me here.