Ball `A` is dropped from the top of a building. At the same instant ball `B` is thrown vertically upwards from the ground. When the balls collide, they are moving in opposite direction and the speed of `A` is twice the speed of `B`. At what fraction of the height of the building did the collision occurs ?

Options

A. `1//3`
B. `2//3`
C. `1//4`
D. `2//5`

Answer

To solve the problem, we need to analyze the motion of both balls A and B and determine the height at which they collide.

### Step-by-Step Solution:

1. **Define Variables:**
- Let the height of the building be \( h \).
- Let the height at which the collision occurs be \( x \).
- The speed of ball A when they collide is \( v_A \).
- The speed of ball B when they collide is \( v_B \).
- According to the problem, \( v_A = 2v_B \).

2. **Motion of Ball A:**
- Ball A is dropped from rest, so its initial velocity \( u_A = 0 \).
- The distance it falls when they collide is \( h - x \).
- Using the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
For ball A:
\[ h - x = 0 \cdot t + \frac{1}{2} g t^2 \implies h - x = \frac{1}{2} g t^2 \]

3. **Motion of Ball B:**
- Ball B is thrown upwards with an initial velocity \( u_B \).
- The distance it travels when they collide is \( x \).
- Using the second equation of motion for ball B:
\[ x = u_B t - \frac{1}{2} g t^2 \]

4. **Relate Speeds at Collision:**
- From the problem statement, we know that:
\[ v_A = 2v_B \]
- The final velocity of ball A can be found using:
\[ v_A = u_A + gt = 0 + gt = gt \]
- The final velocity of ball B is:
\[ v_B = u_B - gt \]
- Thus, substituting into the speed relation:
\[ gt = 2(u_B - gt) \implies gt = 2u_B - 2gt \implies 3gt = 2u_B \implies u_B = \frac{3}{2}gt \]

5. **Substituting \( u_B \):**
- Substitute \( u_B \) back into the equation for \( x \):
\[ x = \left(\frac{3}{2}gt\right)t - \frac{1}{2}gt^2 \]
\[ x = \frac{3}{2}gt^2 - \frac{1}{2}gt^2 = gt^2 \]

6. **Substituting \( x \) into the equation for \( h - x \):**
- We have:
\[ h - x = \frac{1}{2}gt^2 \]
- Thus:
\[ h - gt^2 = \frac{1}{2}gt^2 \implies h = \frac{3}{2}gt^2 \]

7. **Finding the Fraction of Height:**
- Now, we can express the fraction of the height at which the collision occurs:
\[ \frac{x}{h} = \frac{gt^2}{\frac{3}{2}gt^2} = \frac{2}{3} \]

### Final Answer:
The collision occurs at a height that is \( \frac{2}{3} \) of the height of the building.