- #1

Darkmisc

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- Homework Statement
- An elevator descends downwards at 2.5m/s. Seven seconds after this, a rock is thrown upwards at 15m/s. g=9.8m/s^2. When will the rock hit the elevator?

- Relevant Equations
- v = 15 -9.8t.

Hi everyone

I have solutions to this problem, but I am not sure if the premise behind them is correct.

I think this is the reasoning behind the solution for question d.

- Work out how far the elevator travels in the 7s before the rock is thrown, so 7x2.5 = 17.5

- Work out how far the elevator travels in the time it takes for the rock to reach its peak and then fall back to X=0, so 150/49 x 2.5 = 7.65

- This means the elevator will have a head start of 25.15m on the rock when the rock returns to X = 0.

- After this point, the position of the elevator can be given by XL = -25.15 -2.5t

The part of the solution that I'm not sure about is where they give the position of the rock with XR= 15t - 4.9t^2.

It seems like they are starting the clock with XL at -25.15 and XR at 0. However, at this point in time, XR is not starting from rest. It had already fallen for 75/49s. I'm not sure that the equation XR=15t - 4.9t^2 accounts for this.

When the rock returns to X =0, its velocity should be 15 - 150/49*9.8 = -15m/s. Its velocity at subsequent moments should be given by v=-15 -9.8t. Integrating this should give XR= -15t -4.9t^2. I think it is this equation for XR that should be equated with -25.15 -2.5t to get the correct value for t.

Is my reasoning correct? Or are the solutions correct?

Thankshttps://www.physicsforums.com/attachments/313668

https://www.physicsforums.com/attachments/313669

https://www.physicsforums.com/attachments/313671

I have solutions to this problem, but I am not sure if the premise behind them is correct.

I think this is the reasoning behind the solution for question d.

- Work out how far the elevator travels in the 7s before the rock is thrown, so 7x2.5 = 17.5

- Work out how far the elevator travels in the time it takes for the rock to reach its peak and then fall back to X=0, so 150/49 x 2.5 = 7.65

- This means the elevator will have a head start of 25.15m on the rock when the rock returns to X = 0.

- After this point, the position of the elevator can be given by XL = -25.15 -2.5t

The part of the solution that I'm not sure about is where they give the position of the rock with XR= 15t - 4.9t^2.

It seems like they are starting the clock with XL at -25.15 and XR at 0. However, at this point in time, XR is not starting from rest. It had already fallen for 75/49s. I'm not sure that the equation XR=15t - 4.9t^2 accounts for this.

When the rock returns to X =0, its velocity should be 15 - 150/49*9.8 = -15m/s. Its velocity at subsequent moments should be given by v=-15 -9.8t. Integrating this should give XR= -15t -4.9t^2. I think it is this equation for XR that should be equated with -25.15 -2.5t to get the correct value for t.

Is my reasoning correct? Or are the solutions correct?

Thankshttps://www.physicsforums.com/attachments/313668

https://www.physicsforums.com/attachments/313669

https://www.physicsforums.com/attachments/313671