Position, velocity

1. Dec 20, 2005

joshwitkowski

if a person throws a rock strait up with an initial velocity of 25 m/s and a rock is dropped from a height of 50 ft. how do you determine the time at which they will meet in the air?:grumpy:

2. Dec 20, 2005

dicerandom

You find two equations which describe the position of the two rocks as a function of time and then set them equal to one another.

If you have any more questions please post back with whatever work you have attempted so that we have a better idea as to how we can help you.

3. Dec 20, 2005

joshwitkowski

i got x = vi + (at^2)/2 and i plugged in my information and set them equal to eachother but time comes out negative

4. Dec 20, 2005

joex444

Up: $$y_{2} = y_{1} + v_{1y}t + \frac{1}{2}at^2$$ v1 = 25, a = -9.8, what's your initial height when you throw it straight up? from the ground?

Drop: $$y_{2} = y_{1} + v_{1y}t + \frac{1}{2}at^2$$ v1=0, a=-9.8, y1=50.

You could set the y2's equal to each other.

$$50 - \frac{1}{2}gt^2 = 25t - \frac{1}{2}gt^2$$
$$50 = 25t$$

So, t=2 if the rock is thrown up from the ground, and if you take t=2 and put it into each, you end up with the same height, 30.4m.

5. Dec 20, 2005

neurocomp2003

u have 4 motion equations to work from.

6. Dec 22, 2005

daniel_i_l

Maybe it came out negative cause you forgot to put a minus sign on the acceleration of the rock going up?

7. Dec 24, 2005

Morbid Steve

Maybe you should show some work instead of being a lazy person and having others do it for you?