After how long will the balls be at the same height?

[davidw9]

Can somone please walk me through how to do this problem. I've been going at it for wat too long:

A ball is thrown upward from the ground at an inital velocity of 25 m/s; at the same instant, a ball is dropped from rest from a building 15 m high. After how long will the balls be at the same height?

 

[Kurdt]

What have you come up with so far?

 

[davidw9]

i really don't have much at all. I keep trying things that aren't working. I feel like there might should be an x and x-15 for displacement, but I don't know how to apply that.

 

[Kurdt]

What do you know of the kinematic equations? There are a list of them here:

https://www.physicsforums.com/showthread.php?t=110015

What you need to consider are what you know and what you need to find out then. You will need two equations, one for the motion of each ball.

You have:

Ball one v=25m/s

Ball two v=0m/s

ball one x₀=0m

Ball two x₀=15m

acceleration for both = -9.8(m/s)/s

And both are at the same height at the same time.

You should be able to pick what equation you need from that as I've been generous.

 

[davidw9]

ahh, sorry. i know all of that. i know you are trying to find two equations that you can set equal to each other, i just can't figure out how to get to that point of setting them equal.

 

[Kurdt]

Well there are two unknowns, the distance at which they pass which will be equal and the time at which they pass which will be equal. You want to know the time so you set the equations for distance distance equal thus eliminating that term leaving you with one equation and one unknown, and it comes up with a beautifully simple relation.

 

[davidw9]

i keep trying that with the x= v(t)+ 1/2a(t)^2 and keep getting really complicated relationships... ugh.

 

[Kurdt]

OK so using

$$x=x_0+vt+\frac{1}{2}at^2$$

write the equation of motion for both balls and then make them equal to one another. Show me what you get.

 

[davidw9]

wow. i feel stupid. i finally got it. t=.6. thanks a lot.

 

[Kurdt]

No problem. I know we can't give out the answers here but I hope the fact you got it through you own work makes you appreciate it a lot more.