# Homework Help: Motion 2 problems - Tossing hay bales and baseballs

1. Jun 20, 2012

### blackraven

Motion 2 problems -- Tossing hay bales and baseballs

1. The problem statement, all variables and given/known data
1. A hay-baling machine throws each finished bundle of hay 2.0 m up in the air so it can land on a trailer waiting 4.7 m behind the machine.

(a) What must be the speed with which the bundles are launched?

2. A baseball is popped up, remaining aloft for 6.3 s before being caught at a horizontal distance of 83 m from the starting point. What was the launch angle?

2. Relevant equations
1. x = 1/2at^2 +vot +x0
x-xo = d

2. vx = cos*vo
vy = sin*vo

3. The attempt at a solution
1. I used the distance and plugged it into the first equation but it is the wrong answer I believe.
2. I know that in the end I have to divdide vy by vx and take the arctan of that but I do not understand how to find vy and vx. Can someone explain to me step by and step and what that means?

2. Jun 20, 2012

### Saitama

Re: Motion 2 problems -- Tossing hay bales and baseballs

Let's first talk about the vertical motion. At the maximum height, velocity is zero. Initially, the vertical velocity is vsinθ (a component of the initial velocity making an angle θ with the ground). You have initial velocity, final velocity, displacement and the acceleration.
Can you form an equation now?

3. Jun 22, 2012

### xlava

Re: Motion 2 problems -- Tossing hay bales and baseballs

This is going to sound like weird advice, but you're not thinking about the problem right. Don't think "I know that in the end I have to..." Just work through it.

Your equations are sound. Lets start with this: you know the maximum height right? What can you find with that if you know the acceleration of gravity?

4. Jun 22, 2012

### azizlwl

Re: Motion 2 problems -- Tossing hay bales and baseballs

ISEE Method

1. Identify
a) Only one object - the hay bale
b) The object doing 2 works. Going in horizontal and verical motion.

2. Select
y=y0+vy0ty+(1/2)ayty2
x=x0+vx0tx+(1/2)axtx2

Since it involves only one object,
ty=tx

3. Execute
4. Evaluate

Last edited: Jun 23, 2012