Problem about Free Falling Objects

[Ammar w]

Homework Statement

A ball is thrown vertically upward with an initial speed of 20 m/s. Two seconds later, a stone is thrown vertically (from the same initial height as the ball) with an initial speed of 24 m/s. At what height above the release point will the ball and stone pass each other?

Homework Equations

y_f = y_i + v_it + ½ at²

The Attempt at a Solution

The right solution :

The ball is thrown 2 sec before the rock
yball = ½ at² + v_it + y_i
yball = ½(-10)(t+2)² + 20(t+2)
yball = -5t² + 20

yrock = ½ at²+ v_it + y_i
yrock = ½(-10)t² + 24t
yrock = -5t² + 24t

yball = yrock (they pass each other)
-5t² + 20 = -5t² + 24t
t = 5/6 seconds

yrock = -5(5/6)2 + 24(5/6)
yrock = 17 meters

but I don't know why he considered (t) of the ball (t+2) althought the ball is thrown 2 seconds before the stone?
isn't it true if :
t of the ball = t-2
t of the stone = t
?

 

[frogjg2003]

Where is the ball at t=0 and where is the rock at t=0? The reason you use t+2 instead of t-2 is because the equation you used for the rock assumed it was y=0 at t=0, but it was at y=0 at t=-2. t-(-2)=t+2

 

[Ammar w]

Thank you. that helps.
but I will be pleased with more clarification.

 

[frogjg2003]

The equation you used both times, $y=v_0t-\frac{1}{2}gt^2$ is for a projectile starting at position 0m. You have to have t the same for both objects, and the ball is not at its starting position of 0m at t=0s, but it was at t=-2s.

 

[Nickie]

I would like to clarify that the solution provided is correct. The reason for using (t+2) for the ball and (t) for the stone is because the initial height and initial velocity for both objects are the same. This means that the equations for the two objects will have the same form, but with a time difference of 2 seconds. Therefore, to find the point where they pass each other, we need to equate the two equations and solve for time, which gives us t=5/6 seconds.

The reason for using (t+2) for the ball is to account for the 2-second time difference between the ball and the stone. If we used (t-2) for the ball, we would not get the correct solution as it would imply that the ball was thrown 2 seconds after the stone, which is not the case.

In summary, the solution provided is correct and accounts for the time difference between the two objects. It is important to carefully consider the initial conditions and use the appropriate equations to accurately solve the problem.