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. `1:1`
B. `2:1`
C. `3:2`
D. `4:1`

Answer

To solve the problem, we need to find the ratio of the potential energies of two balls projected under different conditions. Let's break down the steps:

### Step 1: Determine the height reached by the first ball
The first ball is projected vertically upwards with an initial speed \( u \). At the highest point of its journey, its final velocity is \( 0 \). We can use the third equation of motion:

\[ v^2 = u^2 + 2as \]

Here, \( v = 0 \) (final velocity), \( a = -g \) (acceleration due to gravity), and \( s = h_1 \) (height reached). Plugging in the values:

\[ 0 = u^2 - 2gh_1 \]

Rearranging gives:

\[ h_1 = \frac{u^2}{2g} \]

### Step 2: Determine the height reached by the second ball
The second ball is projected at an angle of \( 60^\circ \) with the vertical. This means it makes an angle of \( 30^\circ \) with the horizontal (since \( 90^\circ - 60^\circ = 30^\circ \)). The vertical component of the initial velocity for this ball is \( u \cos(30^\circ) \).

Using the formula for the maximum height in projectile motion, we have:

\[ h_2 = \frac{u_y^2}{2g} \]

Where \( u_y = u \cos(30^\circ) = u \cdot \frac{\sqrt{3}}{2} \). Therefore, we can substitute:

\[ h_2 = \frac{(u \cdot \frac{\sqrt{3}}{2})^2}{2g} = \frac{u^2 \cdot \frac{3}{4}}{2g} = \frac{3u^2}{8g} \]

### Step 3: Calculate the ratio of the heights
Now we can find the ratio of the heights \( h_1 \) and \( h_2 \):

\[ \frac{h_1}{h_2} = \frac{\frac{u^2}{2g}}{\frac{3u^2}{8g}} = \frac{u^2 \cdot 8g}{2g \cdot 3u^2} = \frac{8}{6} = \frac{4}{3} \]

### Step 4: Calculate the potential energies
The potential energy \( PE \) at the highest point is given by:

\[ PE = mgh \]

For the first ball:

\[ PE_1 = mg h_1 = mg \cdot \frac{u^2}{2g} = \frac{mu^2}{2} \]

For the second ball:

\[ PE_2 = mg h_2 = mg \cdot \frac{3u^2}{8g} = \frac{3mu^2}{8} \]

### Step 5: Find the ratio of potential energies
Now we find the ratio of the potential energies:

\[ \frac{PE_1}{PE_2} = \frac{\frac{mu^2}{2}}{\frac{3mu^2}{8}} = \frac{mu^2 \cdot 8}{2 \cdot 3mu^2} = \frac{8}{6} = \frac{4}{3} \]

### Final Answer
Thus, the ratio of their potential energies at the highest points of their journey is:

\[ \frac{PE_1}{PE_2} = \frac{4}{3} \]