A particle is dropped from rest from the top of a building of height 100m. At the same instant another particle is projected upward from the bottom of the building What should be the speed of projection of the particle projected from the bottom of building so that the particle cross each other after 1s? Acceleration due to gravity is 10 `m//s^(2)`.

Options

A. 100m/s
B. 95m/s
C. 90m/s
D. 105m/s

Explanation

To solve the problem of two particles moving towards each other, we will follow these steps:

### Step 1: Understand the Problem
We have two particles:
- Particle A is dropped from the top of a building of height 100 m.
- Particle B is projected upwards from the bottom of the building.

We need to find the speed of projection of Particle B such that both particles cross each other after 1 second.

### Step 2: Identify the Given Data
- Height of the building (h) = 100 m
- Time until they cross each other (t) = 1 s
- Acceleration due to gravity (g) = 10 m/s²

### Step 3: Calculate the Position of Particle A after 1 second
Since Particle A is dropped from rest, we can use the second equation of motion:
\[ s = ut + \frac{1}{2}gt^2 \]
Where:
- \( s \) = displacement
- \( u \) = initial velocity (0 m/s, since it is dropped)
- \( g \) = acceleration due to gravity (10 m/s²)
- \( t \) = time (1 s)

Substituting the values:
\[ s_A = 0 \cdot 1 + \frac{1}{2} \cdot 10 \cdot (1)^2 \]
\[ s_A = 0 + \frac{1}{2} \cdot 10 \cdot 1 = 5 \text{ m} \]

So, after 1 second, Particle A has fallen 5 m from the top of the building.

### Step 4: Calculate the Position of Particle B after 1 second
Let the speed of projection of Particle B be \( u \). The distance covered by Particle B in 1 second can also be calculated using the second equation of motion:
\[ s_B = ut - \frac{1}{2}gt^2 \]
Substituting the values:
\[ s_B = u \cdot 1 - \frac{1}{2} \cdot 10 \cdot (1)^2 \]
\[ s_B = u - 5 \text{ m} \]

### Step 5: Set Up the Equation for Crossing
For the two particles to cross each other, the sum of the distances they travel must equal the height of the building:
\[ s_A + s_B = 100 \text{ m} \]
Substituting the expressions for \( s_A \) and \( s_B \):
\[ 5 + (u - 5) = 100 \]
Simplifying the equation:
\[ u = 100 \]

### Step 6: Conclusion
The speed of projection of Particle B should be **100 m/s** for the two particles to cross each other after 1 second.

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