Posted by **Vishnu** on Wednesday, October 8, 2008 at 8:14pm.

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?

- physics -
**bobpursley**, Wednesday, October 8, 2008 at 8:29pm
V2^2 = 2(9.8m/s)(18.8) + 37.1^2

Then, V2 is the square root of v2^2

- physics -
**Vishnu**, Wednesday, October 8, 2008 at 8:36pm
So the answer is 41.8 m?

- physics -
**bobpursley**, Wednesday, October 8, 2008 at 8:52pm
The unit is wrong, of course.

- physics -
**Vishnu**, Wednesday, October 8, 2008 at 8:57pm
41.8 m/s ?

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