A particle is projected vertically up and another is let fall to meet at the same instant. If they have velocities equal in magnitude when they meet, the distance travelled by them are in the ratio of

Options

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

Answer

To solve the problem, we need to analyze the motion of the two particles: one projected vertically upward (let's call it particle A) and the other falling downward (let's call it particle B). We will derive the distances traveled by each particle when they meet, given that their velocities are equal in magnitude at that instant.

### Step-by-Step Solution:

1. **Define Variables:**
- Let the initial velocity of particle A (projected upwards) be \( U \).
- Let the acceleration due to gravity be \( g \).
- Let \( T \) be the time taken for both particles to meet.

2. **Velocity of Particle A:**
- The velocity of particle A when it reaches the meeting point can be expressed as:
\[ V_A = U - gT \]
- Here, \( V_A \) is the upward velocity of particle A at time \( T \).

3. **Velocity of Particle B:**
- The particle B is falling from rest, so its initial velocity \( U_B = 0 \). The velocity of particle B when it reaches the meeting point is:
\[ V_B = gT \]
- Since both particles have equal magnitudes of velocity at the meeting point, we set \( |V_A| = |V_B| \):
\[ U - gT = gT \]

4. **Solve for Initial Velocity:**
- From the equation \( U - gT = gT \), we can rearrange it to find:
\[ U = 2gT \]

5. **Distance Traveled by Particle A:**
- The distance traveled by particle A (upward) can be calculated using the equation of motion:
\[ S_A = UT - \frac{1}{2}gT^2 \]
- Substituting \( U = 2gT \):
\[ S_A = (2gT)T - \frac{1}{2}gT^2 = 2gT^2 - \frac{1}{2}gT^2 = \frac{4gT^2 - gT^2}{2} = \frac{3gT^2}{2} \]

6. **Distance Traveled by Particle B:**
- The distance traveled by particle B (downward) can be calculated as:
\[ S_B = \frac{1}{2}gT^2 \]

7. **Ratio of Distances:**
- Now, we find the ratio of the distances traveled by A and B:
\[ \frac{S_A}{S_B} = \frac{\frac{3gT^2}{2}}{\frac{1}{2}gT^2} = \frac{3gT^2/2}{gT^2/2} = 3 \]
- Thus, the ratio of distances traveled by A and B is:
\[ S_A : S_B = 3 : 1 \]

### Final Answer:
The distance traveled by the two particles is in the ratio of **3:1**.