Kinematic Equation Homework Question

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Homework Help Overview

The discussion revolves around a kinematic problem involving the motion of a rock released from a height of 40 meters. Participants are analyzing the application of kinematic equations to determine the speed of the rock just before it hits the ground, considering gravitational acceleration and displacement.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the use of the kinematic equation v² = v₀² + 2aΔy, questioning the signs of displacement and acceleration. There is also mention of alternative equations for solving the problem, including one that incorporates time.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the signs used in the equations. Some guidance has been offered regarding the direction of acceleration and displacement, but no consensus has been reached on the correct approach.

Contextual Notes

There is a focus on the correct assignment of signs in the context of vertical motion, with some participants suggesting that both displacement and acceleration should be treated as negative when considering downward motion.

vmercadooo
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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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