physics
posted by Vishnu .
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

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 ?