Calculating Collision Time: Astronaut's Experiment on Alien Planet

[Malgrif]

Homework Statement

An Astronaugt is doing an experiment on an alien planet. She fires a ball straight up towards a target ball that is dropped at the same time as the lower ball is launched. If air resistence is ignored, how long does it take for the two balls to collide?

Homework Equations

Kinematics equations.

The Attempt at a Solution

So we know that the two balls have three things in common. Time of collision, acceleration, and the displacement that they'll collide.

d1=.5at^2 and d1=.5at^2 + v1t (for the ball being launched)

I subed the two equations into each other and it didn't really work out... I'm thinking it has something to do with the signs or are my equations wrong all together? Thanks for the help.

 

[LowlyPion]

Your equations are ok, but you need to be careful in setting them up.

If y is positive then v is (+) and g is (-).

Where they collide is going to be y for both as you note and t is the same, so apparently you should end in an answer that depends on initial velocity and the original distance to the target.

So ...

y = v*t - 1/2*g*t²

and

y = d - 1/2*g*t²

 

[Malgrif]

Is there a numerical answer to this question? I used your equations and ended up with t = d/v.

and just to clarify, does your equation mean mean the total distance from the freefalling ball to the ground minus how much the ball free fell before the second ball collided with it?

 

[LowlyPion]

Is there a numerical answer to this question? I used your equations and ended up with t = d/v.

and just to clarify, does your equation mean mean the total distance from the freefalling ball to the ground minus how much the ball free fell before the second ball collided with it?

No.

Yes.

When you set up the equations be careful to consistently express the initial conditions. In the case of the second equation, the target was dropping from height d. So the initial condition for the target, must include its distance above the planet at the start. The height of the one from the surface was 0 of course.

The complete equation applied to both is of the form ...

y = yo + v*t + 1/2*g*t²

... but you need to adjust the signs to reflect which direction is positive y.