# Throwing an object upwards and dropping it

1. May 28, 2014

### Maxo

Two objects with same mass. One is being thrown upwards, and the other one is dropped. Do they have the same speed when hitting ground? I think they have. Let's say the object being thrown upwards goes up 3 meters, then starts accelerating downwards. That will be the same as having dropped the object from 3 meters higher. So one object falls for 3 meters longer than the other. Then it must have more time to accelerate, and so it will have a greater velocity when reaching the ground.

Why is this not correct?

2. May 28, 2014

### Bandersnatch

You managed to explain why pretty clearly why this isn't the case.

3. May 28, 2014

### Maxo

I made a typo, I meant I don't think they have the same speed. But is that correct?

4. May 28, 2014

No,they don't.
This can also be explained using the conservation of energy.
We will throw and drop it at a height of $h$
When you throw one of the object up,you give kinetic energy to it, increasing the total mechanical energy. Now it will have $E=mgh+\frac{1}{2}mv^2$ while the other object had $mgh$ only.
That means that when it hits the ground, the kinetic energy of obj 1 will be greater than that of obj 2

Edit: You guys beat me to it.

5. May 28, 2014

### D H

Staff Emeritus
Why do you think that?

A ball thrown downwards at 10 meters/second will hit the ground with a higher velocity than will a ball that is simply dropped (zero initial velocity). What do you think the difference will be in impact velocity between a ball thrown upwards at 10 meters/second versus one thrown downwards at 10 meters/second?

6. May 28, 2014

### Maxo

Thanks. It's interesting to note that throwing an object up with a certain speed and throwing it downwards with the same speed will eventually hit the ground with the same speed.

7. May 28, 2014

That can also be explained by the conservation of energy.
Throwing down at a speed $v$ at height $h$: $E_M=\frac{1}{2}mv^2+mgh$
Throwing up at that speed $v$ at that height $h$ : $E_M=\frac{1}{2}mv^2+mgh$

I find it easier to use conservation of mechanical energy than using kinematic equations.

8. May 28, 2014

### Maxo

Correct me if I'm wrong, but when using the conservation of mech energy principle in situations involving gravitation one must be careful at only looking at velocity in y-direction. So if there is velocity in x-direction it should not be included in the equation.

9. May 28, 2014

I have only studied velocities in one direction only. So I use mech conservation all the time.

10. May 28, 2014

### CAF123

If you are considering a particle moving in two dimensions along some curve y(x), under the influence of gravity, then relative to a fixed Cartesian coordinate system the total energy is the sum of the particle's kinetic energy and potential energy. That is, $$E_T = \frac{m}{2}\left(\dot{x}^2 + \dot{y}^2\right) + mg \,y(x).$$
Energy is additive, so you can simply add up the kinetic energy in each direction. Under the influence of only conservative forces, this quantity will be fixed, determined by the initial conditions of the particle.

11. May 28, 2014

### Maxo

Wouldn't this give a different energy for an object being thrown horizontally (+x direction) than the same object being dropped? Yet these will have the same velocity when hitting the ground? How is that possible?

12. May 28, 2014

### Staff: Mentor

You are wrong. (But since the x-component of velocity would not change, it doesn't matter.)

13. May 28, 2014

### Staff: Mentor

Why do you think that an object thrown horizontally will have the same velocity upon hitting the ground as a dropped object?

14. May 28, 2014

### A.T.

Related Puzzle:

By accident you walk into an elevator shaft, with the elevator at rest somewhere far below you. Before you start to fall you have just enough time to push one of two buttons:
- UP will instantaneously make the elevator move up with a constant speed V
- DOWN will instantaneously make the elevator move down with a constant speed V
What do you do (to minimize your impact speed) ?

Assume no air resistance, and that when pressing DOWN you will reach the elevator before it reaches ground floor.

15. May 28, 2014

### mattt

Nice one. :-)