I spent all day yesterday trying to figure out how to do these on my own, but I can't seem to conceptualize these two problems.(adsbygoogle = window.adsbygoogle || []).push({});

Problem 1

1. The problem statement, all variables and given/known data

Background

When baseball outfielders throw the ball, they usually allow it to take one bounce on the theory that the ball arrives sooner this way. Suppose that after the bounce the ball rebounds at the same angle θ as it had when released (see figure below) but loses half its speed.

Problem

Assuming the ball is always thrown with the same initial speed, at what angle θ should the ball be thrown in order to go the same distance D with one bounce as one thrown upward at ϕ = 43.8° with no bounce?

2. Relevant equations

My teacher said I needed to use this trig identity:

sin(2u) = 2sin(u)cos(u)

3. The attempt at a solution

I attemped this problem in this manner.

R(1bounce) = (Vi sin2R2R(1 bounce) = R(0 bounce)

sin(2[tex]\Phi[/tex])=2sin([tex]\Theta[/tex])cos([tex]\Theta[/tex])

.4996=sin([tex]\Theta[/tex])cos([tex]\Theta[/tex])

This doesn't seem right. Or at least I don't know the trig identity that can get me [tex]\Theta[/tex] from here.

For this second problem, I figured out the initial height and the range, but I don't know how to work these into a problem that can get the time I need at part C.

1. The problem statement, all variables and given/known data

A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 7.50 m/s at an angle of 16.0° below the horizontal. It strikes the ground 6.00 s later.

(a) How far horizontally from the base of the building does the ball strike the ground? (Xf)

(b) Find the height from which the ball was thrown. (Yi)

(c) How long does it take the ball to reach a point 10.0 m below the level of launching?

Xi = 0 m, Xf = range = 43.26 m

Yi = 188 m, Yi = 0 m

Vxi = 7.21 m/s, Vyi = -2.07 m/s

Vxf = 7.21 m/s, Vyf = -2.07 m/s - 9.8t

Ax = 0 m/s^2, Ay = -9.8 m/s^2

t = 6.00 s

2. Relevant equations

Kinematic equations

3. The attempt at a solution

I initially thought this part (part c) was easy - I just used Yf = Yi + (Vyi)t + (a/2)t^2.

Obviously I didn't account for the change in displacement of X, and treated the problem as if the object was dropped straight down. So I'm stuck, and don't know where to go from here. Any help is greatly appreciated.

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# Homework Help: 2 Two-dimensional motion problems

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