Question

An object is projected with kinetic energy K from point A at an angle 60° with the horizontal. The ratio of the difference in kinetic energies at points B and C to that at point A (see figure), in the absence of air friction is : 

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

Hint:

At the maximum height of a projectile, the kinetic energy is never zero; it is equal to $K \cos^2 \theta$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
In projectile motion, the horizontal component of velocity ($v \cos \theta$) remains constant. Kinetic energy changes as the vertical component of velocity changes with height.

Step 2: Key Formula or Approach:
1. $K_A = \frac{1}{2}mv_0^2 = K$.
2. Horizontal velocity $v_x = v_0 \cos 60^\circ = \frac{v_0}{2}$.
3. At the highest point (Point C), $v_y = 0$, so $K_C = \frac{1}{2}m(\frac{v_0}{2})^2 = \frac{K}{4}$.
Step 3: Detailed Explanation:
If Point B is the point of projection (A) and Point C is the peak:
$K_A = K$.
$K_C = K \cos^2 60^\circ = K (\frac{1}{2})^2 = \frac{K}{4}$.
Difference in K.E. between A and C: $K - \frac{K}{4} = \frac{3K}{4}$.
Ratio to K.E. at A: $\frac{3K/4}{K} = \frac{3}{4}$.
Step 4: Final Answer:
The ratio is 3 : 4.