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: Reason through the vertical velocity component.
Peak height depends only on the upward (vertical) speed, since horizontal motion does not lift the stone. So compare the vertical components of the two launches.
Step 2: First stone, straight up.
Launched vertically with speed $V$, its full speed is vertical: $u_{y1} = V$.
Step 3: Second stone, $60^\circ$ from the vertical.
Its vertical component is $u_{y2} = V\cos 60^\circ = V \times \tfrac{1}{2} = \tfrac{V}{2}$.
Step 4: Heights from vertical speed.
At the top, all vertical kinetic energy converts to potential energy: $h = \dfrac{u_y^2}{2g}$. So $h_1 = \dfrac{V^2}{2g}$ and $h_2 = \dfrac{(V/2)^2}{2g} = \dfrac{V^2}{8g}$.
Step 5: Potential energy ratio.
Equal masses, so $\dfrac{U_1}{U_2} = \dfrac{h_1}{h_2} = \dfrac{V^2/2g}{V^2/8g} = \dfrac{8}{2} = 4$.
Step 6: Conclude.
The ratio of peak potential energies is $4:1$. \[ \boxed{U_1 : U_2 = 4 : 1} \]