Moving block 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

Call the acceleration of gravity g, time of flight t, hor. distance of flight s, release velocity v, potential and kinetic energy PE and KE respectively.
v=s/t
h1=.5*g*t^2 --> t=sqrt(2*h1/g) --> v=s/sqrt(2*h1/g) -->
v^2=s^2/(2*h1/g) = .5*s^2*g/h1
KE at release=.5*m*v^2

If h2 is the height above the ground, the answer is expressed in a quadratic equation. In this problem, the quadratic has no real roots, meaning there is no solution. The height h2 is insufficient to acquire enough energy to travel the distance s from a horizontal release from any height h1 < h2. This is shown below, continuing the above derivation:
PE (converted to KE by going down the slide) = m*g(h2-h1)
KE/(m*g)=PE/(m*g)
.5*v^2/g=h2-h1
.25*s^2/h1=h2-h1
We get a quadratic:
h1^2 - h2*h1 + .25*s^2 = 0
Subtituting 2.6 for h2 and 6.76 for .25*s^2:
h1^2 - 2.6*h1 + 6.76 = 0
In quadratic solution terminology, a=1, b=-2.6 and c=6.76.
The discriminant, b^2-4ac is negative so no real roots.

If h2 is the height above h1, we have a very different solution.
PE (converted to KE by going down the slide) = m*g*h2
KE/(m*g)=PE/(m*g)
.5*v^2/g=h2
.25*s^2/h1=h2
h1=.25*s^2/h2=2.6 m
Plugging this back into the earlier equations,
for release velocity, v^2=5.2^2/(2*2.6/g)=5.2*g;
v=7.141 m/s
for impact velocity vf, vhor=v; t=5.2/v = .7282 s;
vvert=g*t=7.141 m/s
Since vvert=vhor, vf=sqrt(2)*v.