A pellet gun is fired straight downward from the edge of a cliff that is 15 m above the ground. The pallet strickes the ground with a speed of 27 m/s. How far above the cliff edge would the pellet have gone had the gun been fired straight upward ?

Options

A. 4.5 m
B. 22 m
C. 15 m
D. 29 m

Answer

To solve the problem step by step, we will analyze the motion of the pellet in two scenarios: when it is fired downward and when it is fired upward.

### Step 1: Analyze the downward motion
When the pellet is fired downward from the edge of the cliff, it strikes the ground with a speed of 27 m/s after falling a distance of 15 m (the height of the cliff).

Using the equation of motion:
\[ v^2 = u^2 + 2as \]
Where:
- \( v \) = final velocity = 27 m/s (downward)
- \( u \) = initial velocity (unknown)
- \( a \) = acceleration due to gravity = 9.8 m/s² (downward)
- \( s \) = displacement = 15 m (downward)

Substituting the known values into the equation:
\[ 27^2 = u^2 + 2 \cdot 9.8 \cdot 15 \]

Calculating:
\[ 729 = u^2 + 294 \]
\[ u^2 = 729 - 294 \]
\[ u^2 = 435 \]
\[ u = \sqrt{435} \approx 20.84 \, \text{m/s} \]

### Step 2: Analyze the upward motion
Now, we consider the scenario where the pellet is fired upward with the same initial velocity \( u \approx 20.84 \, \text{m/s} \).

Using the same equation of motion:
\[ v^2 = u^2 + 2as \]
Where:
- \( v \) = final velocity = 0 m/s (at the maximum height)
- \( u \) = initial velocity = 20.84 m/s (upward)
- \( a \) = acceleration due to gravity = -9.8 m/s² (upward motion is against gravity)
- \( s \) = displacement (height above the cliff edge, which we want to find)

Substituting the known values into the equation:
\[ 0 = (20.84)^2 + 2 \cdot (-9.8) \cdot s \]

Calculating:
\[ 0 = 435 + (-19.6) \cdot s \]
\[ 19.6s = 435 \]
\[ s = \frac{435}{19.6} \]
\[ s \approx 22.2 \, \text{m} \]

### Step 3: Conclusion
The pellet would have gone approximately 22.2 meters above the cliff edge had the gun been fired straight upward.