Question

A stone is projected vertically upwards with velocity $V$. Another stone of same mass is projected at an angle of $60^\circ$ with the vertical with the same speed ($V$). The ratio of their potential energies at the highest points of their journey, is

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

Hint:

Be extremely careful with wording! An angle of $60^\circ$ with the vertical means the horizontal component uses $\sin(30^\circ)$ or $\cos(60^\circ)$. Since peak vertical height scales with the square of the vertical velocity component ($V_y = V \cos(60^\circ) = \frac{V}{2}$), the height scales down by $\left(\frac{1}{2}\right)^2 = \frac{1}{4}$, leading immediately to a 4 : 1 ratio.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
Two stones of identical mass $m$ are thrown with the same initial velocity magnitude $V$.
The first stone is launched vertically straight up ($90^\circ$ to the horizontal).
The second stone is launched at an angle of $60^\circ$ with respect to the vertical.
We need to determine the ratio of their gravitational potential energies at their respective peak heights.

Step 2: Key Formula or Approach:
The gravitational potential energy at the highest point is directly proportional to the maximum vertical height achieved:
$$U = mgh_{max}$$ The formula for maximum height reached by a projectile launched with speed $u$ at an angle $\theta$ relative to the horizontal is:
$$h_{max} = \frac{u^2 \sin^2\theta}{2g}$$

Step 3: Detailed Explanation:
Let's find the maximum height for both cases.
For the first stone (vertical launch):
The angle with the horizontal is $\theta_1 = 90^\circ$.
$$h_1 = \frac{V^2 \sin^2(90^\circ)}{2g} = \frac{V^2}{2g}$$ For the second stone (angled projectile launch):
The problem states the launch angle is $60^\circ$ with the vertical.
Therefore, its angle with the horizontal is $\theta_2 = 90^\circ - 60^\circ = 30^\circ$.
$$h_2 = \frac{V^2 \sin^2(30^\circ)}{2g} = \frac{V^2 \left(\frac{1}{2}\right)^2}{2g} = \frac{V^2}{8g}$$ Since both stones have identical mass $m$, the ratio of their peak potential energies matches the ratio of their peak heights:
$$\frac{U_1}{U_2} = \frac{mgh_1}{mgh_2} = \frac{h_1}{h_2}$$ $$\frac{U_1}{U_2} = \frac{\frac{V^2}{2g}}{\frac{V^2}{8g}} = \frac{8}{2} = \frac{4}{1}$$

Step 4: Final Answer:
The ratio of their potential energies at the highest points is 4 : 1, which corresponds to option (B).