A ball is projected vertically upwards with a certain initial speed. Another ball of the same mass is projected at an angle of `60^(@)` with the vertical with the same initial speed. At highest points of their journey, the ratio of their potential energies will be

Options

A. (a)`4:1`
B. (b)`3:2`
C. (c)`2:3`
D. (d)`2:1`

Explanation

To solve the problem of finding the ratio of potential energies of two balls projected at different angles, we will follow these steps:

### Step 1: Understand the scenario
We have two balls:
- Ball 1 is projected vertically upwards with an initial speed \( v_0 \).
- Ball 2 is projected at an angle of \( 60^\circ \) with the vertical, also with the same initial speed \( v_0 \).

### Step 2: Determine the maximum height of each ball
The potential energy (PE) of an object at height \( h \) is given by the formula:
\[ PE = mgh \]
where \( m \) is the mass, \( g \) is the acceleration due to gravity, and \( h \) is the height.

For Ball 1 (projected vertically):
- The maximum height \( h_1 \) can be calculated using the formula:
\[ h_1 = \frac{v_0^2}{2g} \]
Thus, the potential energy at the highest point for Ball 1 is:
\[ PE_1 = mg h_1 = mg \left(\frac{v_0^2}{2g}\right) = \frac{mv_0^2}{2} \]

For Ball 2 (projected at \( 60^\circ \)):
- The vertical component of the initial velocity is \( v_{0y} = v_0 \cos(60^\circ) = \frac{v_0}{2} \).
- The maximum height \( h_2 \) for Ball 2 is:
\[ h_2 = \frac{(v_{0y})^2}{2g} = \frac{\left(\frac{v_0}{2}\right)^2}{2g} = \frac{v_0^2}{8g} \]
Thus, the potential energy at the highest point for Ball 2 is:
\[ PE_2 = mg h_2 = mg \left(\frac{v_0^2}{8g}\right) = \frac{mv_0^2}{8} \]

### Step 3: Calculate the ratio of potential energies
Now, we can find the ratio of the potential energies of the two balls:
\[ \text{Ratio} = \frac{PE_1}{PE_2} = \frac{\frac{mv_0^2}{2}}{\frac{mv_0^2}{8}} = \frac{1/2}{1/8} = \frac{1}{2} \times \frac{8}{1} = 4 \]

### Conclusion
The ratio of the potential energies at the highest points of their journeys is:
\[ \text{Ratio} = 4:1 \]