A rock of mass 1.0 kg is dropped from a height of 10 meters.?

Question

What will be its velocity just before it strikes the ground?
What if the height is 100 meters?
*neglect air resistance.

Answer 1

potential energy of the rock=1*9.8*10
=98
kintetic energy just before reaching the ground=98J
v^2=98*2/1
v=undreroot 196

Answer 2

you can also figure it out by applying the conservation of total energy principle; it says, the potential energy at height of 10 meters,
should be equal to the kinetic energy of the rock just before it strikes the ground.
the gravitational potential energy of a mass of M Kg at height h is :
Mgh
and the kinetic potential is :
1/2 MV^2
which V is the velocity of the mass .
thus due to conservation principle we yield :
Mgh=1/2 MV^2 => 2gh=V^2 => V= (2gh)^.5
applying the given parameter yields, V= (2*9.8*10)^.5 = 14 m/s
using h=100 m we get, V=(2*9.8*100)^.5 = 44.2 m/s
the interesting point is the velocity is mass independent and we should only take care of height.

Answer 3

Ek = 0.5 x M x V^2 as a result the two^2 = 4 so the respond must be D) develop kinetic means with the aid of component of. it rather is considering the fact that the mass is stable. Ep=mgh as a result the respond must be B ans the gravity and top are stable so subsequently the main crucial mass has the main expertise means.

Answer 4

Vf^2 = Vo^2 + 2ad

Final velocity squared = (original velocity squared) x 2 x (acceleration of gravity = 9.80 m/s^2) x (distance it falls)

Simply plug your information into this equation to solve for the final velocity.

Answer 5

using the formula
(final speed)^2 = (initial speed)^2 + 2(acceleration)(distance)
final speed = sqrt( 0 + 2*9.8*10))
final speed = 14m/s

apply the formula for 100m distance;
final speed = 44.27m/s

Answer 6

1/2 mv*v = mgh
½ v*v = gh
make v as the subject and substitute.

Answer 7

wieght does not matter all you need to know is g=10 or 9.8 m/sec^2, depends,
it looks like you got a h.w. question,
but this should help you figure it out =]