# Easy for you to solve, need an anwser pls

1. Jan 9, 2004

### sero

If a train is going 35 mph and theres a dude standing in a boxcar and he jumps 2 feet in the air, where will he land? In the same spot, or ahead of or behind the spot he jumped from?

currently some friends and I are debating this. some say the same spot some say behind. I say the same spot due to newtons 1st law of motion, and general relativity.

1. what is the anwser
2. what is the equation to solve it ?

2. Jan 9, 2004

### Staff: Mentor

If he jumps straight up, he will fall straight down...landing on the same spot in the boxcar.
No equations needed, but if you insist, the horizontal distance the jumper travels is:
$$d=v_0t$$
This formula works in the frame of the ground or the boxcar. From the boxcar's viewpoint, $v_0 = 0$; from the viewpoint of someone on the ground, $v_0 = 35{mph}$, the same speed as the boxcar itself.

3. Jan 9, 2004

### sero

cont

thanks Doc

4. Jan 9, 2004

Staff Emeritus
Well there is a caveat. Before the guy jumps he's at rest with respect to the train. This means he shares the same forward speed the train does, relative to the ground, and has a forward momentum of his mass times that speed. When he jumps straight up, his momentum - and therefore his forward velocity - would be conserved if there was no air. And in that case he would still have the same speed as the train and would come down where he started.

But there is air resistance; some of his momentum is transferred to the air, and he loses a litle speed. So when he comes down he hasn't moved quite as far ahead relative to the ground as the train has, and he lands just a little behind his starting spot. Maybe only an inch or to, but iot would be measurable.

I don't think we need to consider coriolis force. Imagine the train is at the equator.

5. Jan 9, 2004

### Staff: Mentor

It's a boxcar, not a flatcar. Assume it's closed and the air is carried along.