Two Stones Thrown Upward: Solving the Problem

[matt23721]

Hey guys, I am 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 can't 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!

 

[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.

 

[matt23721]

ut?? or do you mean at?

 

[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.

 

[matt23721]

ok but how do i find 'u'??

 

[belliott4488]

ok but how do i find 'u'??

You've already said what it is ...

 

[matt23721]

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

 

[belliott4488]

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

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

 

[matt23721]

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

 

[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.