# Question 120e6

Mar 6, 2017

There are two methods by which this can be solved.

1. With the help of Gravitational Potential Energy consideration.
2. With the help of Kinetic energy consideration.

I opt for the GPE method.

a. When the ball is thrown upwards from the ground level, its potential energy is zero and it has maximum kinetic energy.
b. As it rises up, the gravity acting against the direction of motion, its velocity decreasing. Part of its kinetic energy gets converted into its potential energy.
c. As it reaches maximum height, its velocity is momentarily zero. At this point all its kinetic energy has been converted in to its potential energy.

Assuming that no energy is lost against air resistance and applying Law of conservation of energy we have
Initial Kinetic Energy$=$Potential Energy at the maximum height
$= m g h$
$= \frac{50}{1000} \times 9.81 \times 100 = 49.05 J$ (Using SI system of units)

Let ${h}_{1}$ be the height where kinetic energy is reduced to 70%. From Law of Conservation of energy we know that at this height remaining energy, which is 30%, is potential energy of the ball. Setting up the equation we get
mgh_1=mghxx30%#
$\implies {h}_{1} = 100 \times \frac{30}{100} = 30 m$