A 60. kg student jumps from the 10. meter platform at ASU's swimming complex into the pool below.
a. Determine her Eg at the top of the platform.
b. How much Ek does she possess at impact?
What is her velocity at impact?
c. Repeat steps a and b for a 75 kg diver.
d. If she jumped from a platform that was twice as high, how many times greater would be her velocity at impact (compare to 60 kg woman)?
e. How much higher would the platform have to be in order for her velocity to be twice as great (compare to 60 kg woman at 10.m height)?
To solve this problem, we can use the principles of gravitational potential energy, kinetic energy, and velocity.
a) To determine the potential energy at the top of the platform, we can use the formula Eg = mgh, where Eg is the gravitational potential energy, m is the mass, g is the acceleration due to gravity, and h is the height.
Eg = 60 kg * 9.8 m/s^2 * 10 m
Eg = 5880 Joules
Therefore, the student's gravitational potential energy at the top of the platform is 5880 Joules.
b) At the impact, all of the potential energy is converted into kinetic energy. We can use the formula Ek = 0.5 * mv^2 to calculate the kinetic energy, where Ek is the kinetic energy, m is the mass, and v is the velocity.
To calculate the kinetic energy, we need to find the velocity first. We can use the conservation of energy principle:
Eg = Ek
5880 Joules = 0.5 * 60 kg * v^2
v^2 = (2 * 5880 Joules) / 60 kg
v^2 = 196 Joules / kg
v^2 = 3.267 m^2/s^2
Taking the square root of both sides, we get:
v = √3.267 m/s
v ≈ 1.81 m/s
Therefore, the student possesses approximately 3.267 Joules per kilogram of kinetic energy at impact, and her velocity is approximately 1.81 m/s.
c) To repeat the steps for a 75 kg diver, we can use the same formulas but substitute the diver's mass.
a. For the gravitational potential energy:
Eg = mgh
Eg = 75 kg * 9.8 m/s^2 * 10 m
Eg = 7350 Joules
b. For the kinetic energy and velocity:
Ek = 0.5 * mv^2
7350 Joules = 0.5 * 75 kg * v^2
v^2 = (2 * 7350 Joules) / 75 kg
v^2 = 196 Joules / kg
v ≈ 4.43 m/s
Therefore, the 75 kg diver possesses approximately 4.43 Joules per kilogram of kinetic energy at impact, and their velocity is approximately 4.43 m/s.
d) If the student jumps from a platform that is twice as high, we can compare the velocities at impact for the 60 kg student.
For the initial jump with a height of 10 m, we found the velocity to be approximately 1.81 m/s. With a height of twice that (20 m), we can calculate the velocity using the same formulas:
Eg = 60 kg * 9.8 m/s^2 * 20 m
Eg = 11,760 Joules
Ek = 0.5 * 60 kg * v^2
11,760 Joules = 0.5 * 60 kg * v^2
v^2 = (2 * 11,760 Joules) / 60 kg
v^2 = 392 Joules / kg
v = √392 m/s
v ≈ 6.26 m/s
Therefore, the velocity at impact from a platform twice as high would be approximately 6.26 m/s, which is approximately 3.45 times greater than the velocity at impact from a 10 m platform for the 60 kg student.
e) To find the height required for the velocity to be twice as great as the 60 kg woman at a 10 m height, we can set up an equation using the formulas for gravitational potential energy, kinetic energy, and velocity.
Let's call the required height "h":
Eg = Ek
mgh = 0.5 * mv^2
60 kg * 9.8 m/s^2 * h = 0.5 * 60 kg * (2 * v)^2
9.8 * h = 2 * (1.81 m/s)^2
h = 2 * (1.81 m/s)^2 / 9.8 m/s^2
h ≈ 0.66 m
Therefore, the height of the platform would have to be approximately 0.66 m higher for the velocity at impact to be twice as great as the 60 kg woman's velocity at the 10 m height.
To solve these questions, we will need to apply the principles of energy and motion.
a. Determine her Eg at the top of the platform:
Eg (gravitational potential energy) is given by the formula: Eg = m * g * h
Given:
m (mass) = 60 kg
g (acceleration due to gravity) = 9.8 m/s^2
h (height) = 10 m
Substituting the values into the formula:
Eg = 60 kg * 9.8 m/s^2 * 10 m
Eg = 5880 Joules
Therefore, her Eg at the top of the 10-meter platform is 5880 Joules.
b. Determine her Ek (kinetic energy) and velocity at impact:
Ek is given by the formula: Ek = 0.5 * m * v^2
Given:
m (mass) = 60 kg
Ek (kinetic energy) = ?
v (velocity) = ?
At impact, all of her initial potential energy is converted into kinetic energy.
So, Ek at impact is equal to Eg at the top of the platform.
Ek = 5880 Joules
To find v (velocity), we use the formula:
Ek = 0.5 * m * v^2
Substitute the known values:
5880 Joules = 0.5 * 60 kg * v^2
Rearrange the equation to solve for v:
v^2 = (2 * 5880 Joules) / 60 kg
v^2 = 196 m^2/s^2
v = √(196 m^2/s^2)
v = 14 m/s
Therefore, her Ek at impact is 5880 Joules, and her velocity at impact is 14 m/s.
c. Repeat steps a and b for a 75 kg diver:
For the 75 kg diver, we repeat the same steps using the new mass value.
a. Determine the Eg at the top of the platform:
m (mass) = 75 kg
g (acceleration due to gravity) = 9.8 m/s^2
h (height) = 10 m
Eg = 75 kg * 9.8 m/s^2 * 10 m
Eg = 7350 Joules
Therefore, the Eg of the 75 kg diver at the top of the 10-meter platform is 7350 Joules.
b. Determine the Ek (kinetic energy) and velocity at impact:
m (mass) = 75 kg
Ek = Eg [At impact, all of the initial potential energy is converted to kinetic energy.]
Ek = 7350 Joules
Ek = 0.5 * m * v^2
7350 Joules = 0.5 * 75 kg * v^2
v^2 = (2 * 7350 Joules) / 75 kg
v^2 = 196 m^2/s^2
v = √(196 m^2/s^2)
v = 14 m/s
Therefore, the Ek of the 75 kg diver at impact is 7350 Joules, and the velocity at impact is 14 m/s.
d. If she jumped from a platform that was twice as high, how many times greater would be her velocity at impact (compared to the 60 kg woman)?
To solve this, we can use the principle of energy conservation, assuming no energy loss due to air resistance. The initial Eg will be converted entirely into Ek at the bottom while neglecting other factors like friction.
Given:
For 60 kg woman:
Eg = 5880 Joules (from part a)
Ek = 5880 Joules (from part b)
v = 14 m/s (from part b)
For a new height of 20 m:
Eg_new = m * g * h_new
Eg_new = 60 kg * 9.8 m/s^2 * 20 m
Eg_new = 11760 Joules
Using the principle of energy conservation:
Eg + Ek = Eg_new + Ek_new
5880 Joules + 5880 Joules = 11760 Joules + Ek_new
Ek_new = 5880 Joules
To find the velocity at impact (v_new), we can use the formula:
Ek_new = 0.5 * m * v_new^2
5880 Joules = 0.5 * 60 kg * v_new^2
v_new^2 = (2 * 5880 Joules) / 60 kg
v_new^2 = 196 m^2/s^2
v_new = √(196 m^2/s^2)
v_new = 14 m/s
Therefore, the velocity at impact for the 60 kg woman from a 20-meter platform is the same as the velocity at impact for a 10-meter platform, which is 14 m/s.
e. How much higher would the platform have to be in order for her velocity to be twice as great (compared to the 60 kg woman at a 10 m height)?
Using the same principle of energy conservation:
Eg + Ek = Eg_new + Ek_new
5880 Joules + 5880 Joules = Eg_new + 2 * Ek
Solving for Eg_new:
Eg_new = 5880 Joules + 11760 Joules
Eg_new = 17640 Joules
To find the new height (h_new), we can rearrange the formula for Eg:
Eg_new = m * g * h_new
17640 Joules = 60 kg * 9.8 m/s^2 * h_new
h_new = 30 m
Therefore, the platform would have to be 30 meters high for her velocity to be twice as great as the 60 kg woman's velocity from a 10-meter platform.