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 of two stones, we will follow these steps:

### Step 1: Define Variables
Let:
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( t_1 \) = time taken by the first stone after being dropped
- \( t_2 \) = time taken by the second stone after being thrown downwards
- The second stone is thrown 1 second after the first stone, so \( t_2 = t_1 - 1 \)

### Step 2: Distance Travelled by Each Stone
The distance travelled by the first stone (which is dropped) can be calculated using the equation of motion:
\[ s_1 = \frac{1}{2} g t_1^2 \]
For the second stone (which is thrown downwards with an initial velocity of \( 20 \, \text{m/s} \)):
\[ s_2 = u t_2 + \frac{1}{2} g t_2^2 \]
Substituting \( u = 20 \, \text{m/s} \) and \( t_2 = t_1 - 1 \):
\[ s_2 = 20(t_1 - 1) + \frac{1}{2} g (t_1 - 1)^2 \]

### Step 3: Set Distances Equal
Since both stones will have travelled the same distance when the second stone overtakes the first:
\[ s_1 = s_2 \]
Substituting the expressions for \( s_1 \) and \( s_2 \):
\[ \frac{1}{2} g t_1^2 = 20(t_1 - 1) + \frac{1}{2} g (t_1 - 1)^2 \]

### Step 4: Substitute \( g \) and Simplify
Substituting \( g = 10 \):
\[ \frac{1}{2} \cdot 10 t_1^2 = 20(t_1 - 1) + \frac{1}{2} \cdot 10 (t_1 - 1)^2 \]
This simplifies to:
\[ 5 t_1^2 = 20(t_1 - 1) + 5(t_1 - 1)^2 \]

### Step 5: Expand and Rearrange
Expanding the right-hand side:
\[ 5 t_1^2 = 20t_1 - 20 + 5(t_1^2 - 2t_1 + 1) \]
This further simplifies to:
\[ 5 t_1^2 = 20t_1 - 20 + 5t_1^2 - 10t_1 + 5 \]
Combining like terms:
\[ 5 t_1^2 = 10t_1 - 15 + 5t_1^2 \]
Subtracting \( 5t_1^2 \) from both sides gives:
\[ 0 = 10t_1 - 15 \]

### Step 6: Solve for \( t_1 \)
Solving for \( t_1 \):
\[ 10t_1 = 15 \implies t_1 = 1.5 \, \text{s} \]

### Step 7: Calculate Distance
Now, we can find the distance travelled by either stone. Using \( s_1 \):
\[ s_1 = \frac{1}{2} g t_1^2 = \frac{1}{2} \cdot 10 \cdot (1.5)^2 = 5 \cdot 2.25 = 11.25 \, \text{m} \]

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

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