- #1
Gazaueli
- 6
- 0
Free Fall simple question, Answer check ASAP please!
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?
d=\upsilon_{i}t + \frac{1}{2}at^{2}
\upsilon_{f}^{2}= \upsilon_{i}^{2} + 2ad
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!
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!