A projectile is fired from the origin with a velocity `v_(0)` at an angle `theta` with x-axis. The speed of the projectile at an altitude `h` is .

Options

A. a)`v_(0) cos theta`
B. b)`sqrt(v_(0)^(2) - 2gh)`
C. c)`sqrt(v_(0)^(2) sin^(2)theta-2gh)`
D. d)None of these

Explanation

To find the speed of a projectile at a certain altitude \( h \), we can use the principle of conservation of mechanical energy. Here’s a step-by-step solution:

### Step 1: Identify the initial conditions
- The projectile is fired from the origin (0,0) with an initial velocity \( v_0 \) at an angle \( \theta \) with the x-axis.

### Step 2: Write down the initial energy
- At the initial point (point O), the potential energy (PE) is 0 (since it is at ground level) and the kinetic energy (KE) is given by:
\[ KE_0 = \frac{1}{2} m v_0^2 \]
- Therefore, the total mechanical energy at point O is:
\[ E_0 = PE_0 + KE_0 = 0 + \frac{1}{2} m v_0^2 = \frac{1}{2} m v_0^2 \]

### Step 3: Write down the energy at height \( h \)
- At the height \( h \) (point A), the potential energy is:
\[ PE_A = mgh \]
- The kinetic energy at this point (with speed \( v \)) is:
\[ KE_A = \frac{1}{2} m v^2 \]
- Therefore, the total mechanical energy at point A is:
\[ E_A = PE_A + KE_A = mgh + \frac{1}{2} m v^2 \]

### Step 4: Apply the conservation of energy principle
- According to the conservation of mechanical energy:
\[ E_0 = E_A \]
- Substituting the expressions for \( E_0 \) and \( E_A \):
\[ \frac{1}{2} m v_0^2 = mgh + \frac{1}{2} m v^2 \]

### Step 5: Simplify the equation
- Cancel the mass \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{1}{2} v_0^2 = gh + \frac{1}{2} v^2 \]
- Rearranging gives:
\[ \frac{1}{2} v^2 = \frac{1}{2} v_0^2 - gh \]
- Multiplying through by 2:
\[ v^2 = v_0^2 - 2gh \]

### Step 6: Solve for \( v \)
- Taking the square root of both sides:
\[ v = \sqrt{v_0^2 - 2gh} \]

### Conclusion
The speed of the projectile at an altitude \( h \) is given by:
\[ v = \sqrt{v_0^2 - 2gh} \]