A projectile is fired from the top of an 80 m high cliff with an initial speed of `30(m)//(s)` at an unknown angle. The speed when its hits the ground.

Options

A. `50(m)//(s)`
B. `100(m)//(s)`
C. `45(m)//(s)`
D. `20(m)//(s)`

Explanation

To solve the problem of finding the speed of a projectile when it hits the ground after being fired from the top of an 80 m high cliff with an initial speed of 30 m/s, we can apply the principle of conservation of mechanical energy. Here’s the step-by-step solution:

### Step 1: Understand the problem
We have a projectile fired from a height (h) of 80 m with an initial speed (u) of 30 m/s. We need to find the final speed (v) just before it hits the ground.

### Step 2: Identify the energies at the two points
- At point A (the top of the cliff):
- Kinetic Energy (KE_A) = (1/2) * m * u²
- Potential Energy (PE_A) = m * g * h

- At point B (just before hitting the ground):
- Kinetic Energy (KE_B) = (1/2) * m * v²
- Potential Energy (PE_B) = 0 (since we take the ground level as the reference point)

### Step 3: Apply the conservation of mechanical energy
According to the conservation of mechanical energy:
\[ KE_A + PE_A = KE_B + PE_B \]

Substituting the expressions for kinetic and potential energy:
\[ \frac{1}{2} m u^2 + mgh = \frac{1}{2} m v^2 + 0 \]

### Step 4: Simplify the equation
We can cancel the mass (m) from all terms since it appears in every term:
\[ \frac{1}{2} u^2 + gh = \frac{1}{2} v^2 \]

### Step 5: Rearrange the equation to solve for v
Rearranging gives:
\[ \frac{1}{2} v^2 = \frac{1}{2} u^2 + gh \]
Multiplying through by 2:
\[ v^2 = u^2 + 2gh \]

### Step 6: Substitute the known values
- Initial speed, \( u = 30 \, \text{m/s} \)
- Height, \( h = 80 \, \text{m} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

Now substituting these values into the equation:
\[ v^2 = (30)^2 + 2 \cdot 10 \cdot 80 \]
\[ v^2 = 900 + 1600 \]
\[ v^2 = 2500 \]

### Step 7: Calculate the final speed
Taking the square root of both sides:
\[ v = \sqrt{2500} \]
\[ v = 50 \, \text{m/s} \]

### Final Answer
The speed of the projectile when it hits the ground is **50 m/s**.

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