Ball and stone thrown upward - kinematics

[david12]

Here is another question.I try to get the answer but ...

1.A ball is trown vertically upward with an initial speed of 15m/s.then,2.2 s later ,A stone is thrown straight up (from the same initial height as the ball)with an initial speed of 21.3m/s.
the acceleration of g=9.8m/s^2. How far above the release point will the ball and stone pass each other?

Here is what I did.

I use the same time for both (2.2second). the equation is Y=Y(inital) + v(inital)t + 1/2 gt^2

After i got both distance I subtracted. which is gave me the didtance b/n them.

 

[tiny-tim]

solution

Hi david12!

I use the same time for both (2.2second).

ah … that's what you've done wrong.

t is unknown

you have to work out two equations relating y and t … one for the ball and one for the stone …

and it helps to use different letters for them …

say y₁ = f(t₁) and y₂ = f(t₂)

and then you solve for y₁ = y₂ and t₁ = t₂ + 2.2