How A ball is dropped from rest according the diagram below

How A ball is dropped from rest according the diagram below. traveling at the point. A? fast is t raveling just before the How ground at point B? won. work power 22 U power 13m .3m distance Time. time distance Power me.

Solution

There are two problems and the value of the total height in the first problem is not clear ;next time please upload image with more resolution , but we solve the 2 problems assuming the total height is 3.0 m in 1st problem

1) the velocity of the ball after reaching 1.5 m is gained by converting the potential into kinetic energy

            the potential energy converted to kinetic form at 1.5 m can be given by

                        potential enrgy of the ball = m g (3.0 m-1.5 m ) = m v^2 / 2

                                                            9.8 * 1.5 = v ^2 / 2

                                                                   v = 29.4 m/ sec

by the time the ball fall to ground all the gravitational potential energy is converted to kinetic energy

        the potential enrgy of the ball falling from hight = mgh = kinetic energy = mv^2/ 2 , m is mass , v is velocity

                                               from above equation v = sqrt (2gh)

                                                                          v = 7.668 m/sec

2nd problem

Potential energy of the block at a height h = weight * height = mg * h     (weight = mg ) m is mass , g is acceleration due to gravity

                                         Potential energy = 73N*24m = 1752 Joules