SSC IMD SA 2017 – Question 89

When a ball is thrown upwards, at which point is its potential energy maximum?

• At the time of throwing
• On the ground
• Just before hitting the ground
• At the highest point

Detailed Explanation:

The correct answer is d) सबसे ऊंचे स्थान पर (At the highest point).

Principle of Conservation of Mechanical Energy:

When a ball of mass $m$ is projected vertically upward with an initial velocity $u$:

• Gravitational Potential Energy (PE): Given by the formula:
  $$PE = mgh$$
  where $g$ is the acceleration due to gravity and $h$ is the vertical height above the reference ground level.
• Kinetic Energy (KE): Given by $KE = \frac{1}{2}mv^2$.
• At the moment of throw ($h = 0$): The potential energy is minimum (zero), and kinetic energy is at its maximum ($KE = \frac{1}{2}mu^2$).
• During the ascent: As the ball rises, its velocity decreases due to gravitational deceleration. Kinetic energy is progressively converted into potential energy.
• At the highest point ($h = h_{\max}$): The instantaneous velocity of the ball becomes zero ($v = 0$). Therefore, the kinetic energy becomes zero ($KE = 0$), and all mechanical energy is transformed into gravitational potential energy. Thus, potential energy is maximum at the highest point ($PE_{\max} = mgh_{\max}$).