An object is released from rest from a top of a building 90 meters high. Neglect air friction. What is the volume of its acceleration?
Calculate the time it takes to reach the floor?
With what velocity does it reach the floor?
How fast is it moving when it is 50 meter above the floor?

1. a = 9.8 m/s²

2. t = 4.28 s

3. Vf = 42 m/s

4. Vf = 28 m/s  

Explanation:

1.

Since, the body is under free fall motion. Therefore, the value of its acceleration shall be equal to the acceleration due to gravity.

a = 9.8 m/s²

2.

The time taken by the ball to reach the ground can be calculated by using second equation of motion:

h = Vi t + (1/2)gt²

where,

h = height = 90 m

Vi = initial velocity = 0 m/s

t = time taken = ?

Therefore,

90 m = (0 m/s)t + (1/2)(9.8 m/s²)t²

t = √(18.36 s²)

t = 4.28 s

3.

In order to find final velocity we use first equation of motion:

Vf = Vi + gt

Vf = 0 m/s + (9.8 m/s²)(4.28 s)

Vf = 42 m/s

4.

when the ball is at height of 50 m, it means it has covered:

h = 90 m - 50 m = 40 m

we use third equation of motion at this point:

2gh = Vf² - Vi²

(2)(9.8 m/s²)(40 m) = Vf² - (0 m/s)²

Vf = √(784 m²/s²)

Vf = 28 m/s

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer