Answer:
y₁ = 37.2 m, y₂ = 22,6 m
Explanation:
For this exercise we can use the kinematic equations
For the globe, with index 1
y₁ = y₀ + v₁ t
For the shot with index 2
y₂ = 0 + v₂ t - ½ g t²
At the point where the position of the two bodies meet is the same
y₁ = y₂
y₀ + v₁ t = v₂ t - ½ g t²
14 + 8.40t = 27.0 t - ½ 9.8 t²
4.9 t² - 18.6 t + 14 = 0
t² - 3,796 t + 2,857 = 0
Let's look for time by solving the second degree equation
t = [3,796 ±√(3,796 2 - 4 2,857)] / 2
t = [3,796 ± 1,727] / 2
t₁ = 2.7615 s
t₂ = 1.03 s
Now we can calculate the distance for each time
y₁ = v₂ t₁ - ½ g t₁²
y₁ = 27 2.7615 - ½ 9.8 2.7615²
y₁ = 37.2 m
y₂ = v₂ t₂ - ½ g t₂²
y₂ = 27 1.03 - ½ 9.8 1.03²
y₂ = 22,612 m
Your potential energy at the top of the hill was (mass) x (gravity) x (height) .
Your kinetic energy at the bottom of the hill is (1/2) x (mass) x (speed)² .
If there was no loss of energy on the way down, then your kinetic energy
at the bottom will be equal to your potential energy at the top.
(1/2) x (mass) x (speed)² = (mass) x (gravity) x (height)
Divide each side by 'mass' :
(1/2) x (speed)² = (gravity) x (height) . . . The answer we get
will be the same for every skater, fat or skinny, heavy or light.
The skater's mass doesn't appear in the equation any more.
Multiply each side by 2 :
(speed)² = 2 x (gravity) x (height)
Take the square root of each side:
<u>Speed at the bottom = square root of(2 x gravity x height of the hill)</u>
We could go one step further, since we know the acceleration of gravity on Earth:
Speed at the bottom = 4.43 x square root of (height of the hill)
This is interesting, because it says that a hill twice as high won't give you
twice the speed at the bottom. The final speed is only proportional to the
<em>square root </em>of the height, so in order to double your speed, you need to
find a hill that's <em>4 times</em> as high.
<span>From his observations of cannon drilling, Count Rumford concluded that heat could NOT be a form of C. matter.
Although all matter contains heat energy, heat itself is not a form of matter, but just one of its integral parts.
</span>
Since the surface of the sea is already at 1 atm
therefore the pressure due to hydraulics is 4 atm.
We know that hydraulic pressure is:
P = ρ g h
where ρ is density, g is gravity and h is height
P = 1000 kg/m^3 * 9.81 m/s^2 * h = 4 atm * 101325 Pa /
atm
h = 41.34 m = 135.5 feet
