A stone is dropped from the top of a tower of height h=60m. Simultaneously another stone is projected vertically upwards from the foot of the tower. They meet at a height `(2h)/3` from the ground level. The initial velocity of the stone projected upwards is `(g=10ms^(-2))`

Options

A. `20ms^(-1)`
B. `60ms^(-1)`
C. `10ms^(-1)`
D. `30ms^(-1)`

Answer

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

### Step 1: Understand the scenario
We have two stones:
- Stone A is dropped from the top of a tower of height \( h = 60 \, \text{m} \).
- Stone B is projected upwards from the foot of the tower.

They meet at a height \( \frac{2h}{3} \) from the ground, which is \( 40 \, \text{m} \) (since \( h = 60 \, \text{m} \)).

### Step 2: Calculate the distance traveled by each stone
- The distance traveled by Stone A (dropped from the top) when they meet is:
\[ h - \frac{2h}{3} = 60 \, \text{m} - 40 \, \text{m} = 20 \, \text{m} \]
- The distance traveled by Stone B (projected upwards) is:
\[ \frac{2h}{3} = 40 \, \text{m} \]

### Step 3: Analyze the motion of Stone A
For Stone A (dropped from rest):
- Initial velocity \( u_A = 0 \)
- Acceleration \( a_A = g = 10 \, \text{m/s}^2 \)
- Distance traveled \( s_A = 20 \, \text{m} \)

Using the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Substituting the values:
\[ 20 = 0 \cdot t + \frac{1}{2} \cdot 10 \cdot t^2 \]
This simplifies to:
\[ 20 = 5t^2 \quad \Rightarrow \quad t^2 = 4 \quad \Rightarrow \quad t = 2 \, \text{s} \]

### Step 4: Analyze the motion of Stone B
For Stone B (projected upwards):
- Let the initial velocity be \( u_B \) (which we need to find).
- Acceleration \( a_B = -g = -10 \, \text{m/s}^2 \)
- Distance traveled \( s_B = 40 \, \text{m} \)
- Time \( t = 2 \, \text{s} \)

Using the same equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Substituting the values:
\[ 40 = u_B \cdot 2 + \frac{1}{2} \cdot (-10) \cdot (2^2) \]
This simplifies to:
\[ 40 = 2u_B - 20 \]
Rearranging gives:
\[ 2u_B = 40 + 20 \quad \Rightarrow \quad 2u_B = 60 \quad \Rightarrow \quad u_B = 30 \, \text{m/s} \]

### Final Answer
The initial velocity of the stone projected upwards is \( \boxed{30 \, \text{m/s}} \).

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