Kinematic Equation Homework Question

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Gwen releases a rock from a 40 m tower, and the problem involves calculating its final speed using kinematic equations. The initial velocity is 0, acceleration due to gravity is 9.8 m/s², and the displacement is -40 m. The calculation initially yields a negative value for v², leading to confusion about the signs of displacement and acceleration. Clarification indicates that both displacement and acceleration should be treated as negative when considering downward motion, and the correct final speed of the rock is determined to be 28 m/s.
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Homework Statement


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 = ?


Homework Equations


v2 = v02 + 2aΔy


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.
 
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vmercadooo said:

Homework Statement


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 = ?


Homework Equations


v2 = v02 + 2aΔy


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.

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
 
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.
 
vmercadooo said:
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.

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

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.

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).
 
berkeman said:
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

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.
 
zgozvrm said:
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.

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