A ball is dropped from the top of a tower 40 m high. What is its velocity when it has covered 20 m? What would be its velocity when it hits the ground? Take g= 10 m/s²

Given :

Height from which ball is dropped , h = 40 m .

Acceleration due to gravity , g= 10 m/s² .

Initial velocity , u = 0 m/s .

To Find :

Velocity when ball covered 20 m and velocity when it hit the ground .

Solution :

Now , height when ball covered 20 m distance is , 40 - 20 = 20 m .

By equation of motion :

Now , distance covered when body reaches ground is , 40 m .

Putting value h = 40 m in above equation , we get :

Hence , this is the required solution .

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

read more on -