A 48 kg skier starts from rest at a height of H = 28 m above the end of the ski-jump ramp. As the skier leaves the ramp, his velocity makes an angle of 28° with the horizontal. Neglect the effects of air resistance and assume the ramp is frictionless.
(a) What is the maximum height h of his jump above the end of the ramp?
m
(b) If he increased his weight by putting on a backpack, would h then be greater, less or, the same?
To find the maximum height h of the skier's jump, we can use the conservation of mechanical energy.
(a) The initial potential energy of the skier at the top of the ramp is given by mgh, where m is the mass, g is the acceleration due to gravity, and h is the height of the ramp. In this case, m = 48 kg and h = 28 m.
The initial kinetic energy of the skier is assumed to be negligible as he starts from rest. Therefore, the total initial mechanical energy of the skier is given by E_initial = mgh.
The final mechanical energy of the skier at the maximum height h is given by E_final = mgh + 0.5mv^2, where v is the velocity of the skier at the highest point of the jump.
Since there is no air resistance and the ramp is frictionless, the only force acting on the skier is the force due to gravity. This force can be broken down into two components: one parallel to the ramp (mg sinθ) and one perpendicular to the ramp (mg cosθ), where θ is the angle the velocity of the skier makes with the horizontal.
As the skier reaches the highest point of the jump, the perpendicular component of the gravitational force cancels out the centripetal force, leaving only the parallel component of the gravitational force.
Using the equation for force, F = ma, and setting the parallel component of the gravitational force equal to ma (where a is the acceleration), we can solve for a:
mg sinθ = ma
Simplifying, we find a = g sinθ.
Since the acceleration a is constant, we can use the kinematic equation v^2 = u^2 + 2as, where u is the initial velocity (which is 0 in this case), s is the distance traveled, and v is the final velocity.
Substituting the values into the equation, we get:
v^2 = 0 + 2(g sinθ)h
v^2 = 2g sinθh
The maximum height h of the skier's jump occurs when the final velocity v is 0. Therefore, we can set v^2 = 0 and solve for h:
0 = 2g sinθh
h = 0 / (2g sinθ)
Since sin(θ) is positive (since the angle is between 0° and 90°), we can conclude that the maximum height h is 0.
Thus, the maximum height h of the skier's jump above the end of the ramp is 0 m.
(b) Changing the weight of the skier by putting on a backpack would not affect the maximum height h of the jump. The maximum height is determined by the initial energy of the skier and the angle at which he leaves the ramp, both of which do not depend on the weight of the skier.