A drives straight off the edge of a cliff that is 54m high.the police at the scene of the accident obseve that the point of impact 130m from the base of the cliff.How fast was the car traveling when it went over the cliff?
How long to fall 54 meters with no initial vertical velocity?
h = (1/2) g t^2
54 = 4.9 t^2
t = sqrt (54/4.9)
then
130 = u t
so
u = 130/t
To find the speed at which the car was traveling when it went over the cliff, we can use the principle of conservation of energy. The potential energy of the car at the top of the cliff is converted to kinetic energy as it falls.
The potential energy of an object at height h is given by the formula: PE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height.
In this case, we can assume the mass of the car is constant, and the acceleration due to gravity is approximately 9.8 m/s^2.
The kinetic energy of the car just before it hits the ground is equal to the potential energy it had at the cliff's edge, as there is no non-conservative force (like air resistance) acting on the car. So we can equate the potential energy and kinetic energy:
PE = KE
mgh = (1/2)mv^2
Where v is the velocity of the car just before it hits the ground.
Since mass (m) cancels on both sides of the equation, we are left with:
gh = (1/2)v^2
To solve for v, we need to substitute the given values:
h = 54m
g = 9.8 m/s^2
Now we can plug in these values and solve for v:
(9.8 m/s^2)(54m) = (1/2)v^2
529.2 = (1/2)v^2
v^2 = 1058.4
v ≈ 32.54 m/s
Therefore, the car was traveling at approximately 32.54 m/s when it went over the cliff.
To determine the speed at which the car was traveling when it went over the cliff, we can use the principle of conservation of energy.
The potential energy of the car at the top of the cliff is converted into kinetic energy when it reaches the bottom. This relationship can be expressed as:
m * g * h = 0.5 * m * v^2
Where:
m = mass of the car (which cancels out on both sides of the equation)
g = acceleration due to gravity (approximately 9.8 m/s^2)
h = height of the cliff (54m)
v = velocity or speed of the car at the point of impact (what we want to calculate)
Plugging in the values into the equation:
9.8 * 54 = 0.5 * v^2
529.2 = 0.5 * v^2
v^2 = 529.2 / 0.5
v^2 = 1058.4
v ≈ √1058.4
v ≈ 32.53 m/s
Therefore, the car was traveling at approximately 32.53 m/s when it went over the cliff.