A man throws a packet from a tower directly aiming at his friend who is standing at a certain distance from the base which is same as the height of the tower. If the packet is thrown with a speed of `4 ms^(-1)` and it hits the ground midway between the tower base & his friend than the height of the tower is `(g = 10 ms^(-2))`

Options

A. 5m
B. 8m
C. 3.2 m
D. 13m

Answer

To solve the problem step by step, we will use the concepts of projectile motion.

### Step 1: Understand the Problem
We have a tower of height \( H \). A man throws a packet from the top of the tower with a speed of \( 4 \, \text{m/s} \) at an angle of \( 45^\circ \) towards his friend who is standing at a distance equal to the height of the tower from the base. The packet hits the ground midway between the tower and the friend.

### Step 2: Set Up the Geometry
Let the height of the tower be \( H \). The distance from the base of the tower to the friend is also \( H \). Therefore, the total horizontal distance from the tower to the friend is \( H \).

Since the packet hits the ground midway, the horizontal distance from the tower to the point where the packet lands is \( \frac{H}{2} \).

### Step 3: Resolve the Initial Velocity
The initial velocity \( v = 4 \, \text{m/s} \) can be resolved into horizontal and vertical components. Since the angle of projection \( \theta = 45^\circ \):

- Horizontal component \( v_x = v \cos(45^\circ) = 4 \cdot \frac{1}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} \, \text{m/s} \)
- Vertical component \( v_y = v \sin(45^\circ) = 4 \cdot \frac{1}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} \, \text{m/s} \)

### Step 4: Find the Time of Flight
The horizontal distance traveled by the packet when it hits the ground is \( \frac{H}{2} \). Using the formula for horizontal motion (since there is no horizontal acceleration):

\[ \text{Distance} = \text{Speed} \times \text{Time} \]
\[ \frac{H}{2} = v_x \cdot t \]
Substituting \( v_x \):
\[ \frac{H}{2} = 2\sqrt{2} \cdot t \]
From this, we can express time \( t \):
\[ t = \frac{H}{4\sqrt{2}} \quad \text{(Equation 1)} \]

### Step 5: Analyze the Vertical Motion
The vertical motion can be described by the equation:
\[ H = v_y \cdot t + \frac{1}{2} g t^2 \]
Substituting \( v_y \) and \( g = 10 \, \text{m/s}^2 \):
\[ H = 2\sqrt{2} \cdot t + \frac{1}{2} \cdot 10 \cdot t^2 \]
Substituting \( t \) from Equation 1:
\[ H = 2\sqrt{2} \cdot \left(\frac{H}{4\sqrt{2}}\right) + 5 \cdot \left(\frac{H}{4\sqrt{2}}\right)^2 \]
Simplifying the first term:
\[ H = \frac{H}{2} + 5 \cdot \frac{H^2}{32} \]
Multiplying through by 32 to eliminate the fraction:
\[ 32H = 16H + 5H^2 \]
Rearranging gives:
\[ 5H^2 - 16H = 0 \]
Factoring out \( H \):
\[ H(5H - 16) = 0 \]
Thus, \( H = 0 \) or \( H = \frac{16}{5} = 3.2 \, \text{m} \).

### Conclusion
The height of the tower is \( H = 3.2 \, \text{m} \).