Question

From an elevated point 'A', a stone is projected vertically upwards. The velocity of the stone when it reaches a distance 'h' below 'A' is double its velocity when it was at a height 'h' above 'A'. The greatest height attained by the stone above 'A' is

• A. \( \frac{6h}{5} \)
• B. \( \frac{5h}{3} \)
• C. \( 2h \)
• D. \( 7h \)

Hint:

In projectile motion, always use the energy conservation method or kinematic equations to solve for unknown heights or velocities.

Answer 1

The Correct Option is A

Let the velocity of the stone when it is at height \( h \) above \( A \) be \( v \).
At height \( h \) below \( A \), the velocity becomes \( 2v \).
Using the equation of motion: \( v^2 = u^2 + 2gh \), where \( u = 0 \) at the maximum height, we can use the principle of conservation of energy. We calculate the height using the relationship between the velocities at different heights, which gives the greatest height as \( \frac{6h}{5} \).