From what height would a compact car have to be dropped to have the same kinetic energy that it has when being driven at 100 km/h?
100 km/h = 27.78 m/s
V = sqrt (2gH)
H = V^2/(2g) = 39.3 m
I agree with the previous answer, except for the excessive number of significant figures. No more than three are justified
Well, if we're talking about a compact car, I think it would be safer for all involved if we just stick to driving it at 100 km/h rather than dropping it from a height. Let's leave the high-flying stunts to the birds and keep both our cars and our insurance premiums intact, shall we?
To determine the height from which a compact car would have to be dropped to have the same kinetic energy as when it is driven at 100 km/h, we can use the principle of conservation of energy.
First, let's calculate the kinetic energy of the car when driven at 100 km/h.
Step 1: Convert the speed from km/h to m/s:
1 km/h = 1000 m / 3600 s = 0.2778 m/s
Therefore, 100 km/h is equivalent to (100 x 0.2778) = 27.78 m/s
Step 2: Calculate the mass of the compact car.
Let's assume the mass of the car is 1000 kg (of course, this value can vary).
Step 3: Use the kinetic energy formula:
Kinetic Energy = (1/2) x mass x velocity^2
Kinetic Energy = (1/2) x 1000 kg x (27.78 m/s)^2
= 385,000 Joules
So, the car's kinetic energy when driven at 100 km/h is 385,000 Joules.
Step 4: Equate the potential energy due to the car's height with its kinetic energy.
Potential Energy = Kinetic Energy
Potential Energy = mgh ("mgh" represents mass x gravity x height)
mgh = 385,000 J
Step 5: Solve for the height (h):
h = 385,000 J / (1000 kg x 9.8 m/s^2)
h ≈ 39.3 meters
Therefore, a compact car would need to be dropped from approximately 39.3 meters in height to have the same kinetic energy it has when being driven at 100 km/h.
To determine the height from which a compact car would have to be dropped to have the same kinetic energy as when it is being driven at 100 km/h, we need to understand the relationship between kinetic energy and potential energy.
The potential energy of an object at height h is given by the equation: P.E. = mgh, where m is the mass of the object, g is the acceleration due to gravity (approximately 9.8 m/s^2 on Earth), and h is the height.
The kinetic energy of the object moving at a speed v is given by the equation: K.E. = (1/2)mv^2, where m is the mass of the object and v is the velocity.
Since we want to determine the height at which the potential energy is equal to the kinetic energy of the compact car at 100 km/h, we can set the two equations equal to each other.
(1/2)mv^2 = mgh
The mass, m, cancels out on both sides, and we can plug in the known values:
(1/2)(1000 kg)(100,000 m^2/s^2) = (1000 kg)(9.8 m/s^2)(h)
Now we can solve for h:
50,000,000 J = 9,800 kg·m^2/s^2·h
Divide both sides of the equation by 9,800 kg·m/s^2:
50,000,000 J / 9,800 kg·m/s^2 = h
h ≈ 5,102 meters
Therefore, the compact car would need to be dropped from a height of approximately 5,102 meters (or 5.1 kilometers) to have the same kinetic energy as when it is being driven at 100 km/h.
Answer: 39.32746882 meters.
How: KE = 1/2massxvelocity^2 = massxheightxaccelerationofGravity.
Your velocity is 100km/h = 27.777..., so:
1/2(27.777...)^2 x mass = (385.8024691 m^2/s^2)xmass.
(385.8024691m^2/s^2)m = mh(9.81m/s^2).
Mass and all units except meters cancel when you solve for height, h, which is the answer I gave above.