A particle is dropped from a height h and at the same instant another particle is projected vertically up from the ground. They meet when the upper one has descended a height h/3. Find the ratio of their velocities at this instant.

Answer

To solve the problem, we need to analyze the motion of both particles and find the ratio of their velocities when they meet. Let's break it down step by step.

### Step 1: Define the motion of the two particles
- Let particle A be the one dropped from height \( h \).
- Let particle B be the one projected upwards from the ground.

### Step 2: Determine the distance traveled by each particle when they meet
- When they meet, particle A has descended a height of \( \frac{h}{3} \).
- Therefore, the distance A has fallen is \( \frac{h}{3} \), and the distance remaining for A is \( h - \frac{h}{3} = \frac{2h}{3} \).
- At the same time, particle B has traveled upwards a distance of \( \frac{2h}{3} \).

### Step 3: Use kinematic equations to find the velocities
1. **For particle A** (dropped from rest):
- Initial velocity \( u_A = 0 \)
- Displacement \( s_A = \frac{h}{3} \)
- Using the equation \( v_A^2 = u_A^2 + 2g s_A \):
\[ v_A^2 = 0 + 2g \left(\frac{h}{3}\right) = \frac{2gh}{3} \]

2. **For particle B** (projected upwards):
- Let \( u_B \) be the initial velocity of particle B.
- Displacement \( s_B = \frac{2h}{3} \) (upwards)
- Using the equation \( v_B^2 = u_B^2 - 2g s_B \):
\[ v_B^2 = u_B^2 - 2g \left(\frac{2h}{3}\right) = u_B^2 - \frac{4gh}{3} \]

### Step 4: Relate the time of flight for both particles
- Since both particles meet at the same time \( t \):
- For particle A:
\[ s_A = \frac{1}{2} g t^2 \implies \frac{h}{3} = \frac{1}{2} g t^2 \implies t^2 = \frac{2h}{3g} \]

- For particle B:
\[ s_B = u_B t - \frac{1}{2} g t^2 \implies \frac{2h}{3} = u_B t - \frac{1}{2} g t^2 \]
Substituting \( t^2 \):
\[ \frac{2h}{3} = u_B t - \frac{1}{2} g \left(\frac{2h}{3g}\right) = u_B t - \frac{h}{3} \]
Rearranging gives:
\[ u_B t = \frac{2h}{3} + \frac{h}{3} = h \implies u_B = \frac{h}{t} \]

### Step 5: Substitute \( u_B \) back to find \( v_B^2 \)
- Now substitute \( u_B = \frac{h}{t} \) into the equation for \( v_B^2 \):
\[ v_B^2 = \left(\frac{h}{t}\right)^2 - \frac{4gh}{3} \]

### Step 6: Find the ratio of velocities
- Now we have:
\[ \frac{v_A^2}{v_B^2} = \frac{\frac{2gh}{3}}{\left(\frac{h}{t}\right)^2 - \frac{4gh}{3}} \]
- To simplify, we can express both in terms of \( h \) and \( g \).

### Step 7: Calculate the ratio \( \frac{v_A}{v_B} \)
- After simplification, we find:
\[ \frac{v_A}{v_B} = \sqrt{\frac{2gh/3}{(h/t)^2 - 4gh/3}} \]

### Final Step: Solve for the ratio
- After substituting the values and simplifying, we find that the ratio of their velocities at the instant they meet is:
\[ \frac{v_A}{v_B} = 2 \]

### Conclusion
The ratio of the velocities of the two particles when they meet is \( 2:1 \).

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