# Homework Help: Kinematic Equation Homework Question

1. Dec 12, 2011

1. The problem statement, all variables and given/known data
Gwen releases a rock at rest from the top of a 40 m tower. If g = 9.8 m/s2 and air resistance is negligible, what is the speed of the rock as it hits the ground?
v0 = initial velocity = 0
a = acceleration = 9.8
Δy = displacement in y direction = -40
v = final velocity = ?

2. Relevant equations
v2 = v02 + 2aΔy

3. The attempt at a solution
v2 = v02 + 2aΔy
v2 = 0 + 2 (9.8) (-40)
v2 = -784
but v2 can't be a negative number. I'm guessing either delta y is not supposed to be negative, but I thought it should be since the rock is falling, not being thrown upwards, or gravity should be negative, but I thought that when motion is downwards, the acceleration due to gravity should be positive.

2. Dec 12, 2011

### Staff: Mentor

Welcome to the PF.

If you have the y-axis pointing up, so that the displacement is -40m, then be sure to use the correct sign on the acceleration due to gravity. In which direction does that acceleration point?

BTW, I would have used a different equation for this problem, but yours is probably fine. I would have used:

y = y(0) + v(0)*t + 1/2 a*t^2

3. Dec 12, 2011

Spinnor:

That's the answer I got, too, but the correct answer is simply 28 m/s. I think something must be wrong with my axes/alignment.

4. Dec 12, 2011

### Staff: Mentor

Make the change I suggested, and you will get the correct answer.

5. Dec 13, 2011

### zgozvrm

No. With kinematics problems, it's always best to think of vertical motion as positive in the upward direction and negative in the downward direction. Therefore, in this case, both displacement and acceleration are negative (the rock is accelerating downward, and moving in a downward direction).

6. Dec 13, 2011

### zgozvrm

Why would you use that equation? The OP isn't looking for the time. This equation would require more work, whereas the equation used by the OP gives the answer they're looking for directly.

7. Dec 13, 2011

### Staff: Mentor

Good point. The question was asking for the final speed, so the OP equation is the better one to use. I just hardly ever used that equation, that's why I mentioned the other one.

8. Dec 13, 2011

### netgypsy

And I would have used gravitational potential energy at the top equals kinetic energy at the ground, which is very logical.:-)