A rock is tossed straight up with a velocity of 37.1 m/s. When it returns, it falls into a hole 18.8 m deep. What is the rock's velocity as it hits the bottom of the hole?
Responses
The gain in kinetic energy equals the loss of potential energy.
(1/2) M (V2^ - V1^2) = M g *(18.8)
The mass M cancels out and V1 = 37.1 m/s is the initial velcoity
Solve for V2, the final velocity
V2^2 = 2 g*18.8 + V1^2
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My attempt:
V2^2 = 2(9.8m/s)(18.8) + 37.1m/s
= 19.6(18.8) + 37.1
= 368.48 + 37.1
= 405.6 m/s
What am I doing wrong in my calculation?
Responses
physics - drwls, Wednesday, October 8, 2008 at 4:56am
You did not square V1, as required by the formula
physics - Vishnu, Wednesday, October 8, 2008 at 9:26am
The answer is 1744.89 m/s
That's a little fast. Should I Square root this answer?
V2^2 = 2(9.8m/s)(18.8) + 37.1^2
Then, V2 is the square root of v2^2
So the answer is 41.8 m?
The unit is wrong, of course.
41.8 m/s ?
Yes, you should take the square root of the answer you obtained in your calculation. Taking the square root is necessary because the equation you used to solve for V2 gives you the square of the final velocity. Since you are looking for the actual velocity, you need to take the square root.
So, after taking the square root of the result you obtained (405.6 m/s), you will get the correct final velocity.