# Stone in air, (Velocity+acc)

1. Feb 16, 2006

### Emortal

Hi i need help on a question, here goes..

"A stone is thrown vertically upward at a speed of 16m/s at time t=0. A second stone is thrown upward with the same speed 1 second later. At what time are the two stones at the same height?

any hints, suggestions, or exact equations to atk this problem is welcome. Im not good with kinematics, maybe the word problems, i dunno, but i dont like physics in general much..>_>

2. Feb 16, 2006

### Staff: Mentor

What is the equation of motion for the first stone? How long until it reaches the top of its arc?

3. Feb 16, 2006

### Emortal

change in x=vt-1/2at^2

4. Feb 16, 2006

### Emortal

plug in 16m/s and solve for t?. the positions cancel out. leaving -vt=-1/2at^2. -16t=-at^2--> 3.62=t?

5. Feb 16, 2006

### Staff: Mentor

What are you using for the acceleration of gravity?

To help you see what is happening, draw a graph with time on the horizontal and position on the vertical. Draw a parabolic arc starting at the origin and going up and coming back down. That's the plot of the first stone, y1(t). Then plot the second stone y2(t). It will have the same exact arc, right? But just shifted to the right in time by 1 second. Note how the two arcs will cross at some point with the first stone falling past the second stone rising. That's the time t you want to solve for. Then use that time t back in the equation for y1(t) to tell you how high the two stones are at the crossing point.

BTW, since y2(t) is a time shifted version of y1(t), here's a hint that can help you with the equations:

y2(t) = y1(t-1)

Have at it!

6. Feb 17, 2006

### phucnv87

http://photo-origin.tickle.com/image/69/3/5/O/69358753O724300783.jpg [Broken]

Last edited by a moderator: May 2, 2017