# Ball kinematics in one dimension

1. Jul 30, 2009

### Cheddar

1. The problem statement, all variables and given/known data
A ball is thrown upward from the top of a 25.0-m-tall building with an initial speed of 12m/s.
At the same time, a person is running on the ground at a distance of 31.0m from the building.
What is the person's average speed if he catches the ball at the bottom of the building?

2. Relevant equations
vertical displacement = final velocity(squared) - initial velocity(squared) / (2 * acceleration)
time = final velocity - initial velocity / acceleration

3. The attempt at a solution
So the first equation gives me the displacement above the building.
Using the 2nd equation I obtained the amount of time it takes to reach that displacement.
I then multiplied that time by 2 for the time up and back down to initial position.
This is as far as I get.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jul 30, 2009

### kuruman

This is known as a "two object catch-up" problem. All catch-up problems in kinematics are solved using the same procedure. Here it is.
1. Write an expression giving the position of object 1 at any time t.
2. Write an expression giving the position of object 2 at any time t.
3. If we denote the catch-up time by tC, the two expressions must be equal when evaluated at tC. Say this with an equation and solve for tC.
4. Once you have tC, you can find anything else the problem is asking using the appropriate kinematic equation(s).

3. Jul 30, 2009

### Chrisas

What you are attempting to find is the time for the ball to hit the ground. Once you have that, then you can determine how fast the person on the ground has to run to cover the 31 meters in that amount of time.

What you have so far is the time for the ball to travel up and then back down to the roof height of the building. Now you need add the amount of time for the ball to fall from there to the ground. You know the distance it has to fall since you are given the building height. However, it is not a simple ball drop problem because the ball already has an initial velocity as it passes downward past the building roof top. You need to either recognize or calculate what that initial downward velocity is and then find the time for it to hit the ground.

Alternately, instead of multiplying the time you already found by 2. Just use the time it takes for the ball to travel up and momentarily stop. Use that time to find how height the ball is above the ground. From that point, it is a simple ball drop motion with zero initial velocity. Add the two times to get the total time.

4. Jul 30, 2009

### Cheddar

So, the ball hits the ground in 3.793811901 seconds?
31.0 meters/ 3.793811901 seconds = 8.2 m/s (runner's average velocity)?

5. Jul 30, 2009

### Chrisas

I would agree with that answer.

6. Jul 30, 2009

### kuruman

You got it.

7. Jul 30, 2009

### Cheddar

Thank you both. Very well explained.