Question

A man stands on roof of a house and drops a ball of mass M (from rest) from a height of 10 m above the ground. Its speed just before it hits the ground is?

Answer 1

Vf = 14 [m/s]

Step-by-step explanation:

To solve this problem we must use the following equation of kinematics, which is independent of the mass of the object.

`v_(f) ^(2) =v_(o) ^(2) +(2*g*x)`

where:

Vf = final velocity [m/s]

Vo = initial velocity = 0

g = gravity acceleration = 9.81 [m/s²]

x = distance = 10 [m]

Now replacing:

Vf² = 0 + (2*9.81*10)

Vf = 14 [m/s]