A half pipe can be thought of as half of a cylinder with radius of 7.00m. If a skater with mass 75.0 kg, starts with an initial velocity of 5.00 m/s at the top of the ramp, (a), what is his velocity at the very bottom of the ramp? The skater then approaches the other quarter ramp and is launched vertically into the air. How high up does he go? (Assume the ramp is frictionless)
energy at top = (1/2)m v^2 + m g h
= 75 (25/2+9.81*7)
energy at bottom the same but h = 0
so
75 (V^2/2) = 75 (25/2+9.81*7)
(note-- the mass does not matter)
Now if he comes up to the same height on the other side he will be moving up at 5 m/s
(1/2)(25) = 9.81 h
where h is the height above the top of the pipe.
To find the skater's velocity at the bottom of the ramp, we can use the principle of conservation of energy. At the top of the ramp, the skater has gravitational potential energy and no kinetic energy. At the bottom of the ramp, the skater has no gravitational potential energy and only kinetic energy.
The skater's gravitational potential energy at the top of the ramp is given by the formula:
PE = mgh
Where:
m = mass of the skater (75.0 kg)
g = acceleration due to gravity (9.81 m/s²)
h = height of the ramp (which we need to calculate)
Since the ramp is half of a cylinder, the height of the ramp is the same as the radius of the full cylinder (7.00 m).
So, at the top of the ramp, the skater's gravitational potential energy is:
PE_top = (75.0 kg)(9.81 m/s²)(7.00 m)
Now let's find the skater's kinetic energy at the bottom of the ramp using the formula:
KE = (1/2)mv²
Where:
m = mass of the skater (75.0 kg)
v = velocity of the skater at the bottom of the ramp (which we need to calculate)
Since the skater has no gravitational potential energy at the bottom, their kinetic energy is equal to the gravitational potential energy at the top:
PE_top = KE_bottom
Substituting the values, we can solve for the velocity at the bottom of the ramp:
(75.0 kg)(9.81 m/s²)(7.00 m) = (1/2)(75.0 kg)v²
Simplifying the equation:
(539.25 kg·m²/s²) = (37.5 kg)v²
Divide both sides by 37.5 kg to isolate v²:
v² = (539.25 kg·m²/s²) / (37.5 kg)
v² ≈ 14.377 m²/s²
Finally, take the square root of both sides to find v:
v ≈ √(14.377 m²/s²)
v ≈ 3.793 m/s
So the skater's velocity at the very bottom of the ramp is approximately 3.793 m/s.
To find the height the skater reaches after leaving the quarter ramp, we can again use the principle of conservation of energy. Since there is no friction, we can assume that the total mechanical energy (potential energy + kinetic energy) remains constant.
At the highest point of the skater's trajectory, all of their initial kinetic energy will be converted into gravitational potential energy. The equation is:
PE = mgh
Where:
m = mass of the skater (75.0 kg)
g = acceleration due to gravity (9.81 m/s²)
h = maximum height reached by the skater (which we need to calculate)
Using the initial velocity of the skater (5.00 m/s) and the final velocity at the top (0 m/s), we can find h:
Initial kinetic energy (KEi) = Final potential energy (PEf)
(1/2)mv² = mgh
(1/2)(75.0 kg)(5.00 m/s)² = (75.0 kg)(9.81 m/s²)h
Simplifying the equation:
(1/2)(75.0 kg)(25.0 m²/s²) = (75.0 kg)(9.81 m/s²)h
Divide both sides by (75.0 kg)(9.81 m/s²) to isolate h:
h = (1/2)(25.0 m²/s²) / (9.81 m/s²)
h ≈ 12.82 m
Therefore, the skater reaches a maximum height of approximately 12.82 m.