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

 

[baba89]

Thank you for sharing your approach to solving this problem. Your solution method is correct. By using the same time for both objects, you have eliminated the time variable and focused on the distance variable. This is a good approach in kinematics problems where we want to find the distance between two objects at a specific time. Keep up the good work!