Question

A stone is projected at an angle of 30° to the horizontal. The ratio of kinetic energy at the point of projection to the potential energy at the highest point of flight will be:

• A. 1:4
• B. 4:1
• C. 3:4
• D. 4:3

Hint:

Projectile motion problems often require a clear understanding of how kinetic and potential energy are converted as the projectile moves.

Answer 1

The Correct Option is B

Step 1: Analyze energy conservation. At projection: \[ KE_i = \frac{1}{2}mv^2 \] At the highest point: \[ PE_f = mgh \quad {and} \quad KE_f = \frac{1}{2}mv_x^2 \] Where \( v_x = v \cos(30^\circ) \). 
Step 2: Calculate the ratio. \[ v_x = v \cos(30^\circ) = v \cdot \frac{\sqrt{3}}{2} \] \[ KE_f = \frac{1}{2}m(v \cdot \frac{\sqrt{3}}{2})^2 = \frac{3}{8}mv^2 \] \[ {Ratio} = \frac{KE_i}{PE_f} = \frac{\frac{1}{2}mv^2}{mgh} \] Assuming \( h = \frac{v_y^2}{2g} \) and \( v_y = v \sin(30^\circ) = \frac{v}{2} \): \[ h = \frac{(\frac{v}{2})^2}{2g} = \frac{v^2}{8g} \] \[ PE_f = m \cdot \frac{v^2}{8g} \cdot g = \frac{mv^2}{8} \] \[ {Ratio} = \frac{\frac{1}{2}mv^2}{\frac{mv^2}{8}} = 4 \]