Troublesome problem?

Question

A block starts at rest and slides down a fric-
tionless track.
The acceleration of gravity is 9.81 m/s2 :
It leaves the track horizontally, striking the
ground.

a) At what height h1 above the ground is the
block released? Answer in units of m.


Givens: h2(second height)=2.6m
Mass m of block =481g
After leaving the track the block lands 5.2m away from the
point it leaves.

What is the speed of the block when it
leaves the track? Answer in units of m/s

What is the speed of the block when it hits
the ground? Answer in units of m/s.

Answer 1

assuming that h2 is the height that the block leaves the track, then the velocity that it leaves with can be found. Find time it takes the block to strike the ground, with the equation y=y(initial)+v(initial)t+(1/2)at^2
v(initial) is zero, a is 9.81 y(inital) is 2.6, so t=.514 sec

Then with that time, find the velocity bu using the same equation, except replace the y's with x's. x(initial) is zero, x is 5.2 m, a is zero, t is .514 sec, so v(initial)=10.1 m/s.

Using an energy balance, we can find the energy at the h2 by using the potential and kinetic energy equations. Potential = mgh, m=.481kg, g=9.81m/s^2, h=2.6m, so potential energy is 12.168 J. It also has kinetic energy, and the equation is (1/2)mv^2. m is .481kg, v is 10.1m/s, so kinetic energy is 2.429 J. Add them together, and we get the total energy to be 14.597 J.

To find the height of h1, use the potential energy equation. At the top, potential is 14.597 J, because energy is conserved. PE=mgh, 14.597=(.481)(9.81)h, so h=3.093 m

Answer 2

simple, the mass has nothing to do with it, since there's no friction constant, or air resistant.

(5.2m)/(2.6m)=2

therefore:

(2 * 2.6)+h2=h1=7.8m