a brick of mass 2.0kg is dropped from a rest position 5.0m above the ground what is its velocity at a height of 3.0m above the ground [g=9.8meter per second square

V^2 = Vo^2 + 2g*h.

Vo = 0.
g = 9.8 m/s^2.
h = 3 m.
V = ?

7.7

60

9.8

To find the velocity of the brick at a height of 3.0m above the ground, you can use the equations of motion.

The equations of motion for an object in free fall are:
1. v = u + gt
2. s = ut + (1/2)gt^2
3. v^2 = u^2 + 2gs

Where:
- v is the final velocity
- u is the initial velocity (which is 0 m/s since the object is dropped from rest)
- g is the acceleration due to gravity (9.8 m/s^2)
- t is the time taken
- s is the displacement

Given:
- Initial height (s) = 5.0m
- Final height (s') = 3.0m
- Initial velocity (u) = 0 m/s
- Acceleration due to gravity (g) = 9.8 m/s^2

First, we can find the time taken (t) for the brick to fall from 5.0m to 3.0m:
Using equation 2: s = ut + (1/2)gt^2
3.0m = 0t + (1/2)(9.8)(t^2)
3.0 = 4.9t^2
t^2 = 3.0/4.9
t ≈ 0.782s (taking the square root)

Now that we know the time taken, we can find the velocity at a height of 3.0m using equation 1: v = u + gt
v = 0 + (9.8)(0.782)
v ≈ 7.64 m/s

Therefore, the velocity of the brick at a height of 3.0m above the ground is approximately 7.64 m/s.

Where is the answer of that