Question

What will the velocity of a 2 kg ball be just before it hits the ground if it is dropped from 75 m?

Answer 1

We can use the equation for the velocity of an object under free fall:

v^2 = u^2 + 2as

where:

v = final velocity (what we want to find)

u = initial velocity (zero in this case)

a = acceleration due to gravity (9.8 m/s^2)

s = distance (75 m)

Plugging in the values, we get:

• v^2 = 0 + 2(9.8)(75)

• v^2 = 1470

• v = sqrt(1470)

• v = 38.3 m/s (rounded to one decimal place)

Therefore, the velocity of the 2 kg ball just before it hits the ground is approximately 38.3 m/s.

This calculation assumes that there is no air resistance, which is an idealized scenario. In reality, air resistance would slow down the ball and cause it to hit the ground with a lower velocity.