A particle `A` is projected verically upwards. Another indentical particle `B` is projected at an angle of `45^(@)`. Both reach the same height. The ratio of the initial kinetic energy of `A` to that of `B` is `-`

Options

A. `1:2`
B. `2:1`
C. `1:sqrt(2)`
D. `sqrt(2):1`

Answer

To solve the problem step by step, we need to find the ratio of the initial kinetic energy of particle A (projected vertically upward) to that of particle B (projected at an angle of 45 degrees) given that both reach the same maximum height.

### Step 1: Understand the relationship between height and initial velocity
For a particle projected vertically upwards, the maximum height \( h \) can be expressed using the formula:
\[ h = \frac{V_1^2}{2g} \]
where \( V_1 \) is the initial velocity of particle A and \( g \) is the acceleration due to gravity.

### Step 2: Write the expression for the height of particle B
For particle B, which is projected at an angle of 45 degrees, the vertical component of its initial velocity \( V_2 \) can be expressed as:
\[ V_{2y} = V_2 \sin(45^\circ) = \frac{V_2}{\sqrt{2}} \]
The maximum height \( h \) for particle B can be expressed as:
\[ h = \frac{(V_{2y})^2}{2g} = \frac{\left(\frac{V_2}{\sqrt{2}}\right)^2}{2g} = \frac{V_2^2}{4g} \]

### Step 3: Set the heights equal
Since both particles reach the same height, we can set the two height equations equal to each other:
\[ \frac{V_1^2}{2g} = \frac{V_2^2}{4g} \]
We can simplify this equation by multiplying both sides by \( 4g \):
\[ 2V_1^2 = V_2^2 \]
This leads to:
\[ V_1^2 = \frac{V_2^2}{2} \]

### Step 4: Find the ratio of initial kinetic energies
The kinetic energy \( K \) of a particle is given by the formula:
\[ K = \frac{1}{2} m V^2 \]
Thus, the initial kinetic energies of particles A and B can be expressed as:
\[ K_1 = \frac{1}{2} m V_1^2 \quad \text{and} \quad K_2 = \frac{1}{2} m V_2^2 \]
Now, we can find the ratio of the initial kinetic energies:
\[ \frac{K_1}{K_2} = \frac{\frac{1}{2} m V_1^2}{\frac{1}{2} m V_2^2} = \frac{V_1^2}{V_2^2} \]
Substituting the relationship we found earlier:
\[ \frac{K_1}{K_2} = \frac{\frac{V_2^2}{2}}{V_2^2} = \frac{1}{2} \]

### Step 5: Conclusion
Thus, the ratio of the initial kinetic energy of particle A to that of particle B is:
\[ K_1 : K_2 = 1 : 2 \]

### Final Answer
The ratio of the initial kinetic energy of A to that of B is \( 1 : 2 \).

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