A melon is dropped from a height of 39.2 m. After it crosses through half that distance , the acceleration due to gravity is reduced to zero by air drag. With what velocity does the melon hit the ground ?

Answer

To solve the problem of the melon being dropped from a height of 39.2 m, we can break it down step by step.

### Step 1: Understand the Problem
The melon is dropped from a height of 39.2 m. It falls under the influence of gravity until it reaches half the distance (19.6 m), at which point air resistance reduces the acceleration due to gravity to zero. We need to find the velocity of the melon just before it hits the ground.

### Step 2: Calculate the Distance Fallen Before Air Resistance Acts
The total height from which the melon is dropped is 39.2 m. When it crosses half that distance, it has fallen:
\[ S = \frac{39.2}{2} = 19.6 \, \text{m} \]

### Step 3: Use the Kinematic Equation
We will use the kinematic equation to find the velocity of the melon after it has fallen 19.6 m. The equation is:
\[ V^2 = U^2 + 2AS \]
Where:
- \( V \) = final velocity
- \( U \) = initial velocity (which is 0, since it is dropped)
- \( A \) = acceleration (which is \( -9.8 \, \text{m/s}^2 \))
- \( S \) = distance fallen (19.6 m)

### Step 4: Substitute Values into the Equation
Since the initial velocity \( U = 0 \):
\[ V^2 = 0 + 2 \times (-9.8) \times (-19.6) \]

### Step 5: Calculate
Now we calculate:
\[ V^2 = 2 \times 9.8 \times 19.6 \]
\[ V^2 = 2 \times 9.8 \times 19.6 = 384.16 \]
Now, take the square root to find \( V \):
\[ V = \sqrt{384.16} \approx 19.6 \, \text{m/s} \]

### Step 6: Conclusion
The velocity of the melon just before it hits the ground is approximately \( 19.6 \, \text{m/s} \).

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