Question

A body thrown vertically upwards reaches a maximum height \( H \). The ratio of the velocities of the body at heights \( \frac{3H}{4} \) and \( \frac{8H}{9} \) from the ground is:

• A. \( 4:9 \)
• B. \( 27:32 \)
• C. \( 3:2 \)
• D. \( 3:8 \)

Hint:

To find velocity at a specific height in projectile motion, use the energy conservation principle rather than kinematic equations for simplicity.

Answer 1

The Correct Option is C

Step 1: Apply the Energy Conservation Principle Using the principle of conservation of mechanical energy: \[ v^2 = u^2 - 2gh. \] At heights \( h_1 = \frac{3H}{4} \) and \( h_2 = \frac{8H}{9} \): \[ v_1^2 = u^2 - 2g \cdot \frac{3H}{4}, \] \[ v_2^2 = u^2 - 2g \cdot \frac{8H}{9}. \]
Step 2: Calculate the Ratio of Velocities Dividing both equations: \[ \frac{v_1^2}{v_2^2} = \frac{(u^2 - \frac{6gH}{4})}{(u^2 - \frac{16gH}{9})}. \] Taking square root: \[ \frac{v_1}{v_2} = \frac{3}{2}. \] % Final Answer Thus, the correct answer is option (3): \( 3:2 \).

Answer 2

Step 1: Apply the Principle of Energy Conservation
As the particle moves downward, its potential energy converts into kinetic energy. Using the mechanical energy conservation law:
\[ v^2 = u^2 - 2gh, \] where \( u \) is the velocity at the highest point and \( v \) is the velocity at height \( h \).
For the two heights \( h_1 = \frac{3H}{4} \) and \( h_2 = \frac{8H}{9} \), the velocities are:
\[ v_1^2 = u^2 - 2g \times \frac{3H}{4} = u^2 - \frac{3gH}{2}, \] \[ v_2^2 = u^2 - 2g \times \frac{8H}{9} = u^2 - \frac{16gH}{9}. \]
Step 2: Calculate the Velocity Ratio
To find the ratio \( \frac{v_1}{v_2} \), divide the two expressions:
\[ \frac{v_1^2}{v_2^2} = \frac{u^2 - \frac{3gH}{2}}{u^2 - \frac{16gH}{9}}. \] Taking the square root of both sides yields:
\[ \frac{v_1}{v_2} = \sqrt{\frac{u^2 - \frac{3gH}{2}}{u^2 - \frac{16gH}{9}}}. \] Given that the ratio simplifies to \( \frac{3}{2} \), we conclude:
\[ \frac{v_1}{v_2} = \frac{3}{2}. \]
Final Result:
Therefore, the velocity ratio at the two specified heights is:
\[ \boxed{3 : 2}. \]