Free Fall: Find Time & Height for Passing Balls

[azn4lyf89]

A ball is thrown upward with speed V. At the instant the ball is thrown, second ball is dropped from rest at a height H above the thrown ball, but not directly above it. Ignore air resistance. How much time elapses before the balls pass each other? Find a value of H such that the ball thrown upward is at its highest point when the dropped ball passes.
For the first question, do I need to come up with an equation for both balls and set them equal to each other to find the time when they pass each other? It seems as if I would have two unknowns when I do this. And for the second question do I need to set the velocity equal to 0 to find when the ball reaches its highest point?

 

[mezarashi]

For the 1st question, review the kinematics equations again. There is an equation where displacement is expressed as a function of: initial velocity, gravity, and time. By equating this displacement, you can then solve for the time.

For the 2nd question, you are right on!

 

[baba89]

I would approach this problem by first setting up a coordinate system to represent the motion of the two balls. Let's say the thrown ball starts at a height of 0 and the dropped ball starts at a height H. The equations of motion for the two balls can be written as follows:

For the thrown ball:
y = Vt - 0.5gt^2 (where y is the height, V is the initial velocity, g is the acceleration due to gravity, and t is time)

For the dropped ball:
y = H - 0.5gt^2 (where y is the height, H is the initial height, g is the acceleration due to gravity, and t is time)

To find the time when the balls pass each other, we can set these equations equal to each other and solve for t:

Vt - 0.5gt^2 = H - 0.5gt^2
Vt = H
t = H/V

So the time when the balls pass each other is simply H/V.

For the second question, in order for the thrown ball to be at its highest point when the dropped ball passes, the velocity of the thrown ball must be 0 at that time. So we can set the velocity equation for the thrown ball equal to 0 and solve for t:

Vt - 0.5gt^2 = 0
t = V/g

Therefore, the value of H that satisfies this condition is H = V^2/g. This means that the initial height of the dropped ball should be equal to the square of the initial velocity of the thrown ball divided by the acceleration due to gravity.

In summary, by setting up and solving the equations of motion for the two balls, we can find the time when they pass each other and the required initial height for the dropped ball to coincide with the highest point of the thrown ball. It is important to note that these calculations are based on the assumption of no air resistance, and in real-life scenarios, air resistance may play a role in the motion of the balls.