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.
5880 J
b. How much Ek does she possess at impact? 5880 J
What is her velocity at impact? 14 m/s
c. Repeat steps a and b for a 75 kg diver.
7350 J and 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 (compare to 60 kg woman)? 19.8 m/s jumping from twice as high increases the speed by only sqrt(2).
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)? I need help with this one.
I'm also just here to check my answer.
Please check this
To calculate the height of the platform required for the velocity to be twice as great, we can use the principle of conservation of mechanical energy.
The initial potential energy (Eg) at the top of the platform is given by the formula:
Eg = m * g * h
Where:
m = mass (60 kg)
g = acceleration due to gravity (9.8 m/s^2)
h = height of the platform
So, at the height of 10 m:
Eg = 60 * 9.8 * 10
Eg = 5880 J
The final kinetic energy (Ek) at impact is also 5880 J, which means all the potential energy is converted into kinetic energy.
To find the velocity at impact (vf), we can use the equation:
Ek = 0.5 * m * vf^2
Rearranging the equation, we get:
vf = sqrt(2 * Ek / m)
vf = sqrt(2 * 5880 / 60)
vf ≈ 14 m/s
Now, let's calculate for the 75 kg diver:
Eg = 75 * 9.8 * 10
Eg = 7350 J
The velocity at impact would still be 14 m/s.
Next, if the person jumps from a platform that is twice as high (20 m), we can find the velocity at impact by using the equation:
vf = sqrt(2 * Eg / m)
However, since we already know that the velocity at impact is proportional to the square root of the height, we can conclude that:
vf (twice as high) = sqrt(2) * vf (10 m)
Using the calculated velocity at 10 m (14 m/s), we have:
vf (twice as high) = sqrt(2) * 14
vf (twice as high) ≈ 19.8 m/s
Therefore, jumping from a platform that is twice as high increases the velocity at impact by only the square root of 2 (approximately 1.41 times).
Now, let's determine the height of the platform required for the velocity to be twice as great as the 10 m height.
We can use the same formula as before:
vf = sqrt(2 * Eg / m)
Since we want vf to be twice the velocity at 10 m (28 m/s), we can set up the following equation:
28 = sqrt(2 * Eg / m)
Squaring both sides of the equation, we get:
28^2 = 2 * Eg / m
784 = 2 * Eg / m
Rearranging the equation, we have:
h = Eg / (0.5 * m * g)
Substituting the values, we get:
h = 784 / (0.5 * 60 * 9.8)
h ≈ 25.34 m
Therefore, the platform would have to be approximately 25.34 meters high for the velocity to be twice as great as the 60 kg woman at a 10 m height.
Please note that the calculations are based on the assumptions of ideal conditions and do not take into account factors such as air resistance or the diver's technique.
To determine how much higher the platform would have to be in order for the 60 kg student's velocity to be twice as great, we can use the principle of conservation of mechanical energy. The total mechanical energy of the system (student+Earth) remains constant before and after the jump, neglecting air resistance.
Let's assume the initial height of the platform is h and the final velocity at impact is 2v. We can calculate the initial potential energy (Eg) and final kinetic energy (Ek) using the given mass of the student (60 kg), acceleration due to gravity (9.8 m/s^2), and the known velocity at impact for the 60 kg student (14 m/s).
Step 1: Calculate the initial potential energy (Eg) at the top of the platform.
Eg = mgh
Eg = 60 kg * 9.8 m/s^2 * 10 m
Eg = 5880 J
Step 2: Calculate the final kinetic energy (Ek) at impact.
Ek = 0.5mv^2
Ek = 0.5 * 60 kg * (14 m/s)^2
Ek = 5880 J
Now, we need to find the new height (h') required for the 60 kg student to have a final velocity at impact of 2v (which is 2 * 14 m/s = 28 m/s).
Step 3: Calculate the new height (h') of the platform.
Ek = 0.5mv^2
Ek = 0.5 * 60 kg * (28 m/s)^2
Ek = 5880 J
Using the formula for kinetic energy, we can solve for h':
h' = Ek / (0.5 * m * g)
h' = 5880 J / (0.5 * 60 kg * 9.8 m/s^2)
h' ≈ 19 meters
Therefore, the platform would have to be approximately 19 meters high for the 60 kg student's velocity to be twice as great as when jumping from a 10 meter platform.
Please note that this calculation assumes idealized conditions and neglects air resistance. In reality, factors like air resistance and other energy losses can impact the actual results.