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Free Fall simple question, Answer check please

  • Thread starter Gazaueli
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Free Fall simple question, Answer check ASAP please!!!

Homework Statement



A flying squirrel falls from the top of a tree, without initial velocity. It travels the last 2 meters of its fall in 0.2s. What was the height of the tree?


Homework Equations



d=[itex]\upsilon[/itex][itex]_{i}[/itex]t + [itex]\frac{1}{2}[/itex]at[itex]^{2}[/itex]

[itex]\upsilon_{f}[/itex][itex]^{2}[/itex]= [itex]\upsilon[/itex][itex]_{i}[/itex][itex]^{2}[/itex] + 2ad


The Attempt at a Solution



1. I first concentrated on the last 2 meters of the squirrel's fall by using the equation d=[itex]\upsilon[/itex][itex]_{i}[/itex]t + [itex]\frac{1}{2}[/itex]at[itex]^{2}[/itex]. I rearranged it to find the initial velocity: [itex]\upsilon[/itex][itex]_{i}[/itex]=[itex]\frac{d-(0.5)at^{2}}{t}[/itex]

For initial velocity, I got

[itex]\upsilon[/itex][itex]_{i}[/itex]=[itex]\frac{(2m)-(0.5)(-9.8m/s/s)(0.2)^{2}}{(0.2)}[/itex]

[itex]\upsilon[/itex][itex]_{i}[/itex]=10.98m/s

2. Now I focused on the rest of the squirrel's fall before the last 2m.

Here I used the equation: [itex]\upsilon[/itex][itex]_{f}[/itex][itex]^{2}[/itex]=[itex]\upsilon[/itex][itex]_{i}[/itex][itex]^{2}[/itex]+2ad and rearranged it to find d: d=[itex]\frac{\upsilon_{f}^{2}-\upsilon_{i}^{2}}{2a}[/itex]

By substitution I got:

d=[itex]\frac{(10.98m/s)^{2}-(0m/s)^{2}}{2(-9.8m/s/s)}[/itex]= -6.15m as displacement.

The displacement here can also be seen as the height, therefore H=6.15m

I then added 6.15m to the last 2m to get H=8.15m

The problem however is that in the corrections provided by my professor, his answer is that the height of the tree is 6.15m

His work looks like this:
[itex]\upsilon[/itex][itex]_{i}[/itex] (last 2m)= [itex]\frac{2m}{0.2s}[/itex]-[itex]\frac{(9.8m/s/s)(0.2s)}{2}[/itex]= 9.02m/s

Did he forget to square the time here?

And in the second part he then calculates: d=[itex]\frac{(9.02m/s)^{2}}{2(9.8m/s/s)}[/itex]+2m=6.15m

Can someone please check if I missed something, or if perhaps my prof's answer is incorrect?

Thank you!
 

Answers and Replies

  • #2
gneill
Mentor
20,792
2,770

Homework Statement



A flying squirrel falls from the top of a tree, without initial velocity. It travels the last 2 meters of its fall in 0.2s. What was the height of the tree?


Homework Equations



d=[itex]\upsilon[/itex][itex]_{i}[/itex]t + [itex]\frac{1}{2}[/itex]at[itex]^{2}[/itex]

[itex]\upsilon_{f}[/itex][itex]^{2}[/itex]= [itex]\upsilon[/itex][itex]_{i}[/itex][itex]^{2}[/itex] + 2ad


The Attempt at a Solution



1. I first concentrated on the last 2 meters of the squirrel's fall by using the equation d=[itex]\upsilon[/itex][itex]_{i}[/itex]t + [itex]\frac{1}{2}[/itex]at[itex]^{2}[/itex]. I rearranged it to find the initial velocity: [itex]\upsilon[/itex][itex]_{i}[/itex]=[itex]\frac{d-(0.5)at^{2}}{t}[/itex]

For initial velocity, I got

[itex]\upsilon[/itex][itex]_{i}[/itex]=[itex]\frac{(2m)-(0.5)(-9.8m/s/s)(0.2)^{2}}{(0.2)}[/itex]

[itex]\upsilon[/itex][itex]_{i}[/itex]=10.98m/s
Recheck that calculation. The value you've given corresponds to the impact speed at the ground, not the speed at 2m. Could be you've selected the result from a different calculation.
2. Now I focused on the rest of the squirrel's fall before the last 2m.

Here I used the equation: [itex]\upsilon[/itex][itex]_{f}[/itex][itex]^{2}[/itex]=[itex]\upsilon[/itex][itex]_{i}[/itex][itex]^{2}[/itex]+2ad and rearranged it to find d: d=[itex]\frac{\upsilon_{f}^{2}-\upsilon_{i}^{2}}{2a}[/itex]

By substitution I got:

d=[itex]\frac{(10.98m/s)^{2}-(0m/s)^{2}}{2(-9.8m/s/s)}[/itex]= -6.15m as displacement.

The displacement here can also be seen as the height, therefore H=6.15m

I then added 6.15m to the last 2m to get H=8.15m

The problem however is that in the corrections provided by my professor, his answer is that the height of the tree is 6.15m

His work looks like this:
[itex]\upsilon[/itex][itex]_{i}[/itex] (last 2m)= [itex]\frac{2m}{0.2s}[/itex]-[itex]\frac{(9.8m/s/s)(0.2s)}{2}[/itex]= 9.02m/s

Did he forget to square the time here?

And in the second part he then calculates: d=[itex]\frac{(9.02m/s)^{2}}{2(9.8m/s/s)}[/itex]+2m=6.15m

Can someone please check if I missed something, or if perhaps my prof's answer is incorrect?

Thank you!
 

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