A 10 kg block being held at rest above the ground is released. The block begins to fall under only the effect?

Question

A 10 kg block being held at rest above the ground is released. The block begins to fall under only the effect of gravity. The block falls into a 10 m deep well. The block is traveling at 10.4 m/sec just as it reaches the bottom of the well. At what altitude (in m) was the block released?
a) 14.6
b) 10.5
c) 8.6
d) 6.7
e) 5.3
f) 22
g) 27
h) 30

Answer 1

None of the above.

Conservation of Energy

Gravitation Potential Energy = Kinetic Energy

mass * gravity * height = ½ * mass * velocity^2

mass = 10 kg
gravity = 9.8 m / sec^2
velocity = 10.4 m / sec (at the instant before impact)
height = height above grade + 10 meter deep well

Solving for height:

Height = (½ * velocity^2) / gravity
(note that the mass of the item is canceled out of the equation)

Height = (½ * 10.4^2) / 9.8
Height = 54.08 / 9.8 = 5.5 meters

5.5 meters is the total height that the item fell. Since the well is 10 m deep, this implies the block was lowered 4.5 m into the well before it was released.

This presents a problem because the question clearly states the block was being held at rest above ground. Perhaps this test was not performed on earth. A different acceleration due to gravity would make more sense.

Answer 2

mgh = ½mv²

the m's cancel

9.8 (h + 10) = 0.5(10.4)²

h = 0.5(10.4)²/9.8 - 10 = -4.48 m

So the block was held inside of the well 4.48 m under ground level before being released. Unfortunately this isn't a choice.

Answer 3

v^2-u^2=2as
10.4^2-0^2=2*10*s
109.6/20=s
5.3(approx)=s