A 155 m long ramp is to be built for a ski jump. If a skier starting from rest at the top is to have a speed no faster than 23 m/s at the bottom, what should be the maximum angle of inclination?
f=final o=original
vf^2=vo^2+2a(delta x)
23^2=0+2(9.8)delta x
delta x=27.0,m
then it's just a trig problem
sine angle=27.0/155
angle=10.0 degs.
To solve this problem, we can use the principles of mechanical energy conservation. The total mechanical energy of the skier at the top of the ramp will be equal to the total mechanical energy at the bottom of the ramp.
At the top of the ramp, the skier is at rest, so the initial kinetic energy is zero. Therefore, the total mechanical energy at the top is equal to the gravitational potential energy:
\(E_{\text{top}} = mgh\)
Where:
- \(m\) is the mass of the skier
- \(g\) is the acceleration due to gravity, approximately 9.8 m/s²
- \(h\) is the height of the ramp (since the ramp is horizontal, this is the vertical height at the top)
At the bottom of the ramp, the skier has a speed of 23 m/s. The total mechanical energy at the bottom is the sum of the kinetic energy and the gravitational potential energy:
\(E_{\text{bottom}} = \frac{1}{2}mv^2 + mgh\)
Where:
- \(m\) is the mass of the skier
- \(v\) is the velocity (speed) of the skier at the bottom
- \(g\) is the acceleration due to gravity
- \(h\) is the height of the ramp (this is the same as at the top, since the ramp is horizontal)
The maximum angle of inclination can be found by equating the mechanical energies at the top and the bottom of the ramp:
\(E_{\text{top}} = E_{\text{bottom}}\)
\(mgh = \frac{1}{2}mv^2 + mgh\)
Canceling out the mass, we can simplify the equation:
\(gh = \frac{1}{2}v^2\)
Solving for the maximum angle \(θ\), we can use the trigonometric relationship:
\(h = l \sin(θ)\)
Where:
- \(l\) is the length of the ramp
- \(θ\) is the angle of inclination
Plugging this into the equation for \(gh\), we get:
\(gl \sin(θ) = \frac{1}{2}v^2\)
Simplifying further, we can solve for \(θ\):
\(\sin(θ) = \frac{1}{2}\frac{v^2}{gl}\)
Finally, to find the maximum angle of inclination \(θ\), we can take the inverse sine:
\(θ = \sin^{-1}(\frac{1}{2}\frac{v^2}{gl})\)
Plugging in the given values:
- \(l = 155\) m (length of the ramp)
- \(v = 23\) m/s (maximum speed at the bottom)
- \(g = 9.8\) m/s² (acceleration due to gravity)
We can now calculate the maximum angle of inclination for the ski jump.