Olympic diver performs a dive from the 10 m platform and achieves a speed of 14 m/s just before entering the water. Neglecting air friction, approximately how many times larger the diver’s speed than the diver would have achieved by performing from a 1 m platform? How much more potential energy does the diver have from the 10 m platform than on a 1 m platform?
the potential energy is 10 times as much
... PE = m g h
the kinetic energy is also 10 times, but velocity is squared in KE
10 times KE means √10 times velocity
To calculate the ratio of the diver's speed when diving from the 10 m platform to the speed when diving from a 1 m platform, we can use the principle of conservation of energy.
The potential energy of an object at height h is given by the equation:
PE = mgh
Where m is the mass of the object, g is the acceleration due to gravity, and h is the height.
Since the diver's speed just before entering the water is given as 14 m/s, we can determine the height from which the diver is diving using the equation for kinetic energy:
KE = 1/2mv^2
Where KE is the kinetic energy, m is the mass of the object, and v is the velocity.
By equating the initial potential energy (at the top of the platform) to the final kinetic energy (just before entering the water), we can solve for the height h.
So, 1/2mv^2 = mgh
Simplifying, we find:
h = v^2 / (2g)
Now let's plug in the values:
For diving from the 10 m platform:
h_10 = (14 m/s)^2 / (2 * 9.8 m/s^2)
= 98 / 19.6
= 5 m
For diving from the 1 m platform:
h_1 = (v_1)^2 / (2 * 9.8 m/s^2)
= (v_1)^2 / 19.6
Now, let's compare the heights:
The ratio of the heights is given by:
Ratio = h_10 / h_1
= 5 m / [(v_1)^2 / 19.6]
Since the ratio depends on the value of v_1, we need the speed of the diver when performing from the 1 m platform to calculate it.
However, we can determine the ratio of the potential energies between the 10 m and 1 m platforms.
The potential energy is given by the equation:
PE = mgh
So, the potential energy difference between the 10 m and 1 m platforms is:
ΔPE = mgh_10 - mgh_1
= mg(h_10 - h_1)
= mg(5 m - h_1)
Now, we need the value of h_1 to calculate the difference in potential energy.
Could you please provide the speed of the diver when diving from the 1 m platform?
To determine the ratio of the diver's speed from a 10 m platform to that from a 1 m platform, we can apply the principles of conservation of energy.
Let's start by assuming the diver's initial height on both platforms is the same, indicating that the only difference is the potential energy at the start. The potential energy (PE) is given by the equation PE = mgh, where m represents the mass of the diver, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height.
On a 10 m platform, the potential energy can be calculated as PE_10m = m * g * 10.
On a 1 m platform, the potential energy can be calculated as PE_1m = m * g * 1.
The difference in potential energy (ΔPE) between the two platforms is:
ΔPE = PE_10m - PE_1m
= m * g * 10 - m * g * 1
= m * g * (10 - 1)
= m * g * 9
Therefore, the potential energy difference from the 10 m platform to the 1 m platform is 9 times larger.
Now, let's consider the diver's speed just before entering the water. When the diver starts from rest, the total energy consists of potential energy at the start and kinetic energy at the end. According to the principle of conservation of energy, the total energy is conserved.
The total energy (E) is given by the equation E = PE + KE, where KE represents the kinetic energy.
On a 10 m platform, the total energy can be written as:
E_10m = PE_10m + KE_10m
= m * g * 10 + 0 (assuming the initial speed is zero)
On a 1 m platform, the total energy can be written as:
E_1m = PE_1m + KE_1m
= m * g * 1 + 0 (assuming the initial speed is zero)
Since the total energy is conserved, we have:
E_10m = E_1m
m * g * 10 = m * g * 1 + KE_1m
KE_1m = m * g * 9
Therefore, the kinetic energy on the 1 m platform is 9 times smaller than the kinetic energy on the 10 m platform.
Since kinetic energy is directly related to the square of the speed (KE = 1/2 * m * v^2), we can conclude that the speed of the diver from the 10 m platform is approximately √9 times larger than the speed from the 1 m platform.
√9 = 3
Hence, the diver's speed from the 10 m platform is roughly 3 times larger than the speed achieved from a 1 m platform. Additionally, the potential energy from the 10 m platform is 9 times greater than the potential energy from the 1 m platform.