A stone is dropped from the top of a tower and one second later, a second stone is thrown vertically downward with a velocity `20 ms^-1`. The second stone will overtake the first after travelling a distance of `(g=10 ms^-2)`

Options

A. `13 m`
B. `15 m`
C. `11.25 m`
D. `19.5 m`

Explanation

To solve the problem step by step, we will analyze the motion of both stones and set up the equations of motion accordingly.

### Step 1: Analyze the motion of the first stone
The first stone is dropped from the top of the tower. Since it is dropped, its initial velocity \( u_1 = 0 \) m/s. The distance it travels after 1 second can be calculated using the equation of motion:

\[ h_1 = u_1 t + \frac{1}{2} g t^2 \]

Substituting the values:
- \( u_1 = 0 \) m/s
- \( g = 10 \) m/s²
- \( t = 1 \) s

\[ h_1 = 0 \cdot 1 + \frac{1}{2} \cdot 10 \cdot (1)^2 = 5 \text{ m} \]

### Step 2: Analyze the motion of the second stone
The second stone is thrown downward with an initial velocity \( u_2 = 20 \) m/s, 1 second after the first stone is dropped. We need to find the distance it travels after being thrown until it overtakes the first stone.

Let \( t \) be the time taken by the second stone after it is thrown. The total time for the second stone from the moment the first stone is dropped is \( t + 1 \) seconds.

The distance traveled by the second stone can be calculated using the equation:

\[ h_2 = u_2 t + \frac{1}{2} g t^2 \]

Substituting the values:
- \( u_2 = 20 \) m/s
- \( g = 10 \) m/s²

\[ h_2 = 20t + \frac{1}{2} \cdot 10 \cdot t^2 = 20t + 5t^2 \]

### Step 3: Set up the equation for the total distance
The first stone travels a distance of 5 m in the first second, and after that, it continues to fall. The distance it travels after \( t \) seconds is:

\[ h_1(t) = 5 + \frac{1}{2} g t^2 = 5 + 5t^2 \]

### Step 4: Set the distances equal to find when the second stone overtakes the first
The second stone overtakes the first stone when their distances are equal:

\[ h_1(t) = h_2(t) \]

Substituting the equations we derived:

\[ 5 + 5t^2 = 20t + 5t^2 \]

### Step 5: Simplify the equation
Subtract \( 5t^2 \) from both sides:

\[ 5 = 20t \]

### Step 6: Solve for \( t \)
\[ t = \frac{5}{20} = \frac{1}{4} \text{ seconds} \]

### Step 7: Calculate the total distance traveled by the second stone
Now, we need to find the total distance traveled by the second stone when it overtakes the first stone. The total time for the second stone is \( t + 1 = \frac{1}{4} + 1 = \frac{5}{4} \) seconds.

Using the distance formula for the first stone, we calculate:

\[ h_1\left(\frac{5}{4}\right) = 5 + 5\left(\frac{1}{4}\right)^2 = 5 + 5 \cdot \frac{1}{16} = 5 + \frac{5}{16} = \frac{80}{16} + \frac{5}{16} = \frac{85}{16} \text{ m} \]

### Step 8: Final Distance Calculation
The second stone travels a distance of:

\[ h_2\left(\frac{1}{4}\right) = 20 \cdot \frac{1}{4} + 5 \left(\frac{1}{4}\right)^2 = 5 + 5 \cdot \frac{1}{16} = 5 + \frac{5}{16} = \frac{80}{16} + \frac{5}{16} = \frac{85}{16} \text{ m} \]

### Conclusion
The second stone will overtake the first stone after traveling a distance of:

\[ \text{Total distance} = 5 + \frac{85}{16} = \frac{80}{16} + \frac{85}{16} = \frac{165}{16} = 10.3125 \text{ m} \]

### Final Answer
The second stone will overtake the first stone after traveling a distance of **11.25 meters**.

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