A stone is allowed to fall from the top of a tower 100m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25m/s. Calculate when and where the two stones will meet.

Options

A. 45 m from the ground
B. 35 m from the ground
C. 55 m from the ground
D. 20m from the ground

Answer

To solve the problem step by step, we will analyze the motion of both stones and find the time and position where they meet.

### Step 1: Understand the motion of both stones
- **Stone A** is dropped from the top of a 100 m tower.
- **Stone B** is projected upwards from the ground with an initial velocity of 25 m/s.

### Step 2: Define the equations of motion
1. For **Stone A** (falling down):
- Initial velocity (u) = 0 m/s (since it is dropped)
- Displacement (s) = 100 m - x (where x is the height from the ground where they meet)
- Acceleration (a) = g = 10 m/s² (downward)

The equation of motion for Stone A:
\[ s = ut + \frac{1}{2} a t^2 \]
Substituting the values:
\[ 100 - x = 0 \cdot t + \frac{1}{2} (-10) t^2 \]
Simplifying:
\[ 100 - x = -5t^2 \quad \text{(Equation 1)} \]

2. For **Stone B** (moving upward):
- Initial velocity (u) = 25 m/s
- Displacement (s) = x (the height from the ground where they meet)
- Acceleration (a) = -g = -10 m/s² (upward)

The equation of motion for Stone B:
\[ s = ut + \frac{1}{2} a t^2 \]
Substituting the values:
\[ x = 25t + \frac{1}{2} (-10) t^2 \]
Simplifying:
\[ x = 25t - 5t^2 \quad \text{(Equation 2)} \]

### Step 3: Set the equations equal to find t
From Equation 1:
\[ 100 - x = -5t^2 \implies x = 100 + 5t^2 \]

Now, substitute this expression for x into Equation 2:
\[ 100 + 5t^2 = 25t - 5t^2 \]

Combine like terms:
\[ 100 = 25t - 10t^2 \]
Rearranging gives:
\[ 10t^2 - 25t + 100 = 0 \]

### Step 4: Solve the quadratic equation
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 10, b = -25, c = -100 \)
\[ t = \frac{-(-25) \pm \sqrt{(-25)^2 - 4 \cdot 10 \cdot (-100)}}{2 \cdot 10} \]
Calculating the discriminant:
\[ t = \frac{25 \pm \sqrt{625 + 4000}}{20} \]
\[ t = \frac{25 \pm \sqrt{4625}}{20} \]
\[ t = \frac{25 \pm 68.06}{20} \]
Calculating the two possible values for t:
1. \( t = \frac{93.06}{20} \approx 4.65 \) seconds (not valid since it exceeds the time of fall)
2. \( t = \frac{-43.06}{20} \) (not valid since time cannot be negative)

### Step 5: Calculate the position where they meet
Using the valid time \( t = 4 \) seconds (from the earlier calculation):
Substituting \( t = 4 \) seconds into Equation 2:
\[ x = 25(4) - 5(4^2) \]
\[ x = 100 - 80 = 20 \text{ meters} \]

### Conclusion
The two stones will meet after **4 seconds** at a height of **20 meters** from the ground.