Question

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

Answer 1

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 :

$v^2=u^2+2gh\\\\v=\sqrt{2\times 10\times 20}\ m/s\\\\v=20\ m/s$

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

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

$v=\sqrt{2\times 10\times 40}\ m/s\\\\v=20\sqrt{2}\ m/s$

Hence , this is the required solution .