A body is projetced from of a top of a tower . Speed with which the body hits the ground

Options

A. Must be dependent on angle of projection
B. Must be dependent on speed of projection
C. Must be greater than speed of projection
D. May be less than speed of projection depending on direction of projection .

Explanation

To find the speed with which a body hits the ground when projected from the top of a tower, we can use the principle of conservation of energy. Here’s a step-by-step solution:

### Step 1: Understand the problem
A body is projected from the top of a tower with an initial speed \( u \) at an angle \( \theta \) to the horizontal. We need to determine the speed \( v \) with which the body hits the ground.

### Step 2: Identify the energies involved
When the body is projected, it has two forms of energy:
1. Kinetic energy due to its initial speed: \( KE_i = \frac{1}{2} m u^2 \)
2. Potential energy due to its height above the ground: \( PE = mgh \)

When the body hits the ground, all potential energy will have converted into kinetic energy. The final kinetic energy when it hits the ground will be:
\[ KE_f = \frac{1}{2} m v^2 \]

### Step 3: Apply the conservation of energy principle
According to the conservation of energy:
\[ KE_i + PE = KE_f \]
Substituting the expressions for kinetic and potential energy, we get:
\[ \frac{1}{2} m u^2 + mgh = \frac{1}{2} m v^2 \]

### Step 4: Simplify the equation
We can cancel \( m \) from all terms (assuming \( m \neq 0 \)):
\[ \frac{1}{2} u^2 + gh = \frac{1}{2} v^2 \]

### Step 5: Rearrange to solve for \( v^2 \)
Multiply the entire equation by 2 to eliminate the fraction:
\[ u^2 + 2gh = v^2 \]

### Step 6: Solve for \( v \)
Taking the square root of both sides gives:
\[ v = \sqrt{u^2 + 2gh} \]

### Conclusion
The speed with which the body hits the ground is given by:
\[ v = \sqrt{u^2 + 2gh} \]