Question

what is the speed of a 330kg beam dropped from 40m​

Answer 1

Answer:

28 m/s

Explanation:

We can use this constant acceleration kinematic equation to find the speed of this beam as it is dropped from 40 m.

Since the object is dropped vertically downwards, we know that the object is in free-fall.

The acceleration of this object is -g, which is -9.8 m/s² since we are making the positive direction upwards and the negative direction downwards.

The initial velocity is 0 m/s since the object is dropped and starts from rest.

The final velocity is what we are trying to solve for.

The displacement in the y-direction is 40 m.

Now we can see which equation contains these 4 variables:

• v² = v₀² + 2aΔx

Plug in the known variables.

• v² = (0)² + 2(-9.8)(-40)

Simplify this equation.

• v² = 784

Take the square root of both sides.

• v = 28 m/s

The speed of the 330 kg beam dropped from 40 m is 28 m/s.

Answer 2

Answer:

28 m/s

Explanation:

Given information :

mass = 330kg

height =40m

v = √2gh

Where ;

g = acceleration due to gravity = 9.8 or 10 m/s

h = height

how much total energy of the beam just as it hits the ground =

mgh

$330\times 10\times 40\\\\\\=132000$

Speed at which  the beam will hit the ground :

$v = \sqrt{2\times 10\times40} \\\\=\sqrt{800}\\\\=20\sqrt{2}\\\\v=28.28427$