Question

I solid iron ball of mass 770kg is used on a building site. The ball is suspended by a rope from a crane. The distance from the point of suspension to the centre of the mass of the ball is 12 m.The ball is pulled back from the vertical and then realeased, it falls through a vertical height of 1.6m and strikes the wall: What is the speed of the ball just before impact?

Answer 1

This is a simple problem of conservation of energy. From the point of release, the potential energy due to vertical height of the ball is converted into kinetic energy.

$mgh = \frac{mv^{2} }{2}$
=> $gh = \frac{v^{2} }{2}$
=> $(9.8).(1.6) = \frac{v^{2} }{2}$
=> $v^{2} = 2.(15.68)$
=> $v = \sqrt{31.36}$
=> $v = 5.6 m/s$

Answer 2

Answer:

WHEN WE PULL THE BALL THEN ENERGY STORED IS MGH= 770INTO 9.8 INTO 1.6.

THIS VALUE YOU CAN EQUATE WITH K.E= 1/2 M V V

FROM HERE YOU WILL GET V NEARLY EQUAL TO 5M/S

Explanation: