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A while ago I had a discussion with someone about the speed of an object when it hits the ground. When does an object hit the ground faster:

  • When you launch the object to the sky so it can fall down;
  • Or when you launch the object to the ground so it doesn't have to deal with much air resistance.

The attraction “The Booster” has an arm that has a constant speed (about 120 km/h) is about 80M high and undergoes full revolutions.

enter image description here

In this situation the cart gets disattached at 90 degrees.

enter image description here

I think that the cart hits the ground with a higher speed when it's launched to the ground because of:

  • The air resistance, when you launch it to the sky it has a longer path to travel, so it needs to travel through much more air.
  • The cart has the same initial speed, when you travel a shorter path there will be less to slow the cart down;
  • Nothing provides the cart with additional energy;
  • I guess that the "final gravity" is the same.
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up vote 2 down vote accepted

In my answer I use Conservation of Energy. Our goal is to maximize the kinetic energy of the cart when it hits the ground. $KE = \frac{1}{2} m v^2$ , so maximizing KE would maximize the velocity (since mass is constant)

First scenario: launching upwards. When the cart travels upwards to its maximum height and then falls back to its initial height, it loses some energy to air resistance. Let's call this energy loss $ \Delta E_1$. When the cart falls from that initial height to the ground, it loses some more energy to air resistance. Let's call this energy loss $\Delta E_2$. The total energy loss is $\Delta E_1 + \Delta E_2$ .

Second scenario: launching downwards When the cart travels downwards, it loses energy $\Delta E_2$ to air resistance.The total energy loss is $\Delta E_2$.

Analysis: $\Delta E_1 < 0$ and $\Delta E_2 < 0$ since when we lose energy, then the change in energy is negative. For example, if you initially had $5 J$ of energy and then loss 2J of energy so you ended up with 3J of energy, then the change of energy is $E_{final} - E_{initial} = 3J - 5J = -2J$ and $-2J < 0$.

In the first scenario, we lost more energy than in the second scenario. Therefore, in the first scenario, we ended up with less energy than in the second scenario. Therefore, there was more energy in the second scenario than in the first, which means the cart hit the ground faster in the second scenario.

When I use Energy in my calculations I refer to the Total Energy that the cart has. This is represented by $E_{total} = KE + PE$ where KE is the kinetic energy which increases with speed and PE is the gravitational potential energy which increases with height.

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Thanks, as far as I know there is nothing more that could make a difference in this case. – Laurence Dec 28 '12 at 7:37

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