# Height question

1. Jan 14, 2008

### matt23721

Hey guys, im new here, i was wondering if you could help me solve the following problems
ok...A stone is thrown vertically upward at a speed of 48.70 m/s at time t=0. A second stone is thrown upward with the same speed 4.390 seconds later. At what time are the two stones at the same height?
now i know the height where they meet is: 7.159s, but i cant solve the other part:
At what height do the two stones pass each other?
What is the downward speed of the first stone as they pass each other?

any help would be appreciated
thanks!

2. Jan 14, 2008

### dst

Use s = ut + 1/2at^2 and solve the first.
As for the second, you calculate a maximum point for y then work from then on.

Just try and solve for a function of the displacement of each one and then differentiate to get velocity results.

3. Jan 14, 2008

### matt23721

ut?? or do you mean at?

4. Jan 14, 2008

### mgb_phys

In these equations 'u' is initial velocity.
'v' is final velocity, 's' is displacement, 't' is time and 'a' is acceleration.

Another equation that is useful for the second part is:
v^2 = u^2 + 2 a s

Be careful to get the signs of u,v and a correct.

5. Jan 14, 2008

### matt23721

ok but how do i find 'u'??

6. Jan 14, 2008

### belliott4488

You've already said what it is ...

7. Jan 14, 2008

### matt23721

but once it reaches the peak... it fall down with accelration = 9.8m/s^2 right?

8. Jan 14, 2008

### belliott4488

Be careful with your signs. If "+" is up, then the grav. accel. is negative. Otherwise, yes, the acceleration due to gravity applies consistently throughout.

9. Jan 14, 2008

### matt23721

wow im so lost, lol... this is pathetic... is there anyway i can compute a parabola?

10. Jan 14, 2008

### belliott4488

Look at the equation for height (s) in Post #2: it gives s as a 2nd order polynomial in t, i.e. a parabola. How's that? Think about that and maybe try plotting it (height as a function of time). Once you can do that for the two stones, you should be well on your way.