Question

A projectile is projected with speed $v$ at an angle $60^\circ$ with the horizontal. Find the ratio of the difference of kinetic energies at point $C$ (at ground) and point $B$ (at highest point) with the kinetic energy at point $C$ as shown in the diagram:

• A. $3:4$
• B. $1:3$
• C. $1:2$
• D. $1:12$

Hint:

At the highest point of a projectile, only horizontal velocity remains. Use $v\cos\theta$ directly to compute kinetic energy at the top.

Answer 1

The Correct Option is A

Concept: In projectile motion (neglecting air resistance):
Mechanical energy is conserved
At the highest point, vertical component of velocity becomes zero
Kinetic energy depends on the square of speed
Step 1: Kinetic energy at point $C$ (ground level) Speed at point $C$ is equal to initial speed $v$. \[ K_C = \frac{1}{2}mv^2 \]
Step 2: Kinetic energy at point $B$ (highest point) At the highest point, vertical velocity is zero. Only horizontal component exists: \[ v_B = v\cos60^\circ = \frac{v}{2} \] \[ K_B = \frac{1}{2}m\left(\frac{v}{2}\right)^2 = \frac{1}{8}mv^2 \]
Step 3: Difference of kinetic energies \[ K_C - K_B = \frac{1}{2}mv^2 - \frac{1}{8}mv^2 = \frac{3}{8}mv^2 \] Step 4: Required ratio \[ \frac{K_C - K_B}{K_C} = \frac{\frac{3}{8}mv^2}{\frac{1}{2}mv^2} = \frac{3}{4} \]
Step 5: Hence, the required ratio is: \[ \boxed{3:4} \]