An object falls freely from a height of 25m onto a roof of a building 5m high. Calculate the velocity with which the object strikes the roof
Assuming 25 m above gnd.
d = 25-5 = 20 m.
V^2 = Vo^2 + 2g*d = 0 + 19.6*20 = 392.
V =
an object falls freely from 5m high the ground, then: the velocity of the object when it reaches the ground is
Well, let's see. An object falling from a height onto a surface can be analyzed using basic physics equations. The formula we need is the velocity equation:
v^2 = u^2 + 2as
Where:
v = final velocity (which is what we want to find)
u = initial velocity (which is zero since the object is dropped)
a = acceleration due to gravity (approximately 9.8 m/s^2)
s = displacement (which is the total height the object falls)
In this case, the displacement (s) is the sum of the height of the building (5m) and the initial height (25m). So, s = 5m + 25m = 30m.
Plugging these values into the equation, we have:
v^2 = 0 + 2 * 9.8 m/s^2 * 30m
Simplifying, we get:
v^2 = 2 * 9.8 m/s^2 * 30m
v^2 = 588 m^2/s^2
Taking the square root of both sides:
v ≈ 24.2 m/s
So, the velocity with which the object strikes the roof is approximately 24.2 m/s. Now, wasn't that a hair-raising calculation?
To calculate the velocity with which the object strikes the roof, we can use the principle of conservation of energy. The potential energy at the starting height will be converted into kinetic energy just before hitting the roof.
The potential energy (PE) of an object at a certain height can be calculated using the formula: PE = m * g * h
Where m is the mass of the object, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height of the object above the ground.
In this case, the starting height (h1) is 25m, and the height of the building (h2) is 5m. The potential energy at the starting height will be equal to the kinetic energy just before hitting the roof.
Since both potential energy and kinetic energy are given by the formula: KE = 0.5 * m * v² (where v is the velocity), we can equate the potential energy at the starting height to the kinetic energy just before hitting the roof:
m * g * h1 = 0.5 * m * v²
Simplifying the equation, we can cancel out the mass (m) on both sides:
g * h1 = 0.5 * v²
Now we can solve for v:
v² = (2 * g * h1)
v = √(2 * g * h1)
Plugging in the values in this case:
v = √(2 * 9.8 * 25)
Solving this equation will give us the velocity with which the object strikes the roof.