Question

A man drops a ball off the edge of a water tower 60 meters tall. How fast is the ball moving before it hits the ground? The acceleration due to gravity is 9.81 m/s².

Answer 1

Given:-

• Height of building , h = 60m

• Final velocity ,v = 0m/s

• Acceleration due to gravity ,g = 9.81m/s²

To Find:-

• Initial velocity,u

Solution:-

By using 3rd equation of motion

• v² = u² +2ah

Substitute the value we get

→ 0² = u² + 2× (-9.81)×60

→ 0 = u² + (-1176)

→ -u²= -1176

→ u² = 1176

→ u = √1176

→ u = 34.29 m/s

∴ The initial velocity of the ball is 34.29m/s.