From what height should a car be dropped to gain the same kinetic energy it would have if it were moving horizontally at 95 km/hr?
v^2 = 2gh
To determine the height from which a car should be dropped to gain the same kinetic energy it would have when moving horizontally at a certain speed, we can use the principle of conservation of energy.
The kinetic energy (KE) of an object can be calculated using the equation: KE = (1/2) * m * v^2, where m represents the mass of the object and v represents its velocity.
In this case, we know the desired velocity of the car, which is 95 km/hr. However, we need to convert it to meters per second (m/s) to match the unit of the kinetic energy equation.
1 km/hr = 1000 m / 3600 s = (5/18) m/s
Therefore, the velocity of the car would be:
v = 95 km/hr * (5/18) m/s = 26.38 m/s (approximately)
Now, we need to find the height from which the car should be dropped to acquire this kinetic energy.
The potential energy (PE) of an object at a certain height can be determined using the equation: PE = m * g * h, where g represents the acceleration due to gravity (approximately 9.8 m/s^2) and h represents the height.
To set the potential energy equal to the desired kinetic energy, we have:
PE = KE
m * g * h = (1/2) * m * v^2
The mass of the car cancels out from both sides of the equation, leaving us with:
g * h = (1/2) * v^2
Now we can solve for h:
h = (1/2) * v^2 / g
Plugging in the values:
h = (1/2) * (26.38 m/s)^2 / 9.8 m/s^2
Calculating this equation gives us the height from which the car should be dropped to acquire the same kinetic energy:
h ≈ 35.01 meters
Therefore, the car should be dropped from approximately 35.01 meters height to gain the same kinetic energy it would have when moving horizontally at 95 km/hr.