Free Fall simple question, Answer check please

[Gazaueli]

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=$\upsilon$$_{i}$t + $\frac{1}{2}$at$^{2}$

$\upsilon_{f}$$^{2}$= $\upsilon$$_{i}$$^{2}$ + 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=$\upsilon$$_{i}$t + $\frac{1}{2}$at$^{2}$. I rearranged it to find the initial velocity: $\upsilon$$_{i}$=$\frac{d-(0.5)at^{2}}{t}$

For initial velocity, I got

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

$\upsilon$$_{i}$=10.98m/s

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

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

By substitution I got:

d=$\frac{(10.98m/s)^{2}-(0m/s)^{2}}{2(-9.8m/s/s)}$= -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:
$\upsilon$$_{i}$ (last 2m)= $\frac{2m}{0.2s}$-$\frac{(9.8m/s/s)(0.2s)}{2}$= 9.02m/s

Did he forget to square the time here?

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

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

Thank you!

 

[gneill]

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=$\upsilon$$_{i}$t + $\frac{1}{2}$at$^{2}$

$\upsilon_{f}$$^{2}$= $\upsilon$$_{i}$$^{2}$ + 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=$\upsilon$$_{i}$t + $\frac{1}{2}$at$^{2}$. I rearranged it to find the initial velocity: $\upsilon$$_{i}$=$\frac{d-(0.5)at^{2}}{t}$

For initial velocity, I got

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

$\upsilon$$_{i}$=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: $\upsilon$$_{f}$$^{2}$=$\upsilon$$_{i}$$^{2}$+2ad and rearranged it to find d: d=$\frac{\upsilon_{f}^{2}-\upsilon_{i}^{2}}{2a}$

By substitution I got:

d=$\frac{(10.98m/s)^{2}-(0m/s)^{2}}{2(-9.8m/s/s)}$= -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:
$\upsilon$$_{i}$ (last 2m)= $\frac{2m}{0.2s}$-$\frac{(9.8m/s/s)(0.2s)}{2}$= 9.02m/s

Did he forget to square the time here?

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

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

Thank you!