A projectile is thrown from a point in a horizontal plane such that the horizontal and vertical velocities are `9.8 ms^(-1)` and `19.6 ms^(-1)`. It will strike the plane after covering distance of

Options

A. `39.2 m`
B. `19.6m`
C. `9.8m`
D. `4.9m`

Explanation

To solve the problem, we will use the formula for the range of a projectile when it is thrown with both horizontal and vertical components of velocity. Here are the steps to find the distance covered by the projectile:

### Step 1: Identify the given values
- Horizontal velocity (ux) = 9.8 m/s
- Vertical velocity (uy) = 19.6 m/s
- Acceleration due to gravity (g) = 9.8 m/s²

### Step 2: Use the range formula for projectile motion
The range (R) of a projectile can be calculated using the formula:
\[ R = \frac{2 \cdot ux \cdot uy}{g} \]

### Step 3: Substitute the values into the formula
Now, substitute the known values into the range formula:
\[ R = \frac{2 \cdot 9.8 \cdot 19.6}{9.8} \]

### Step 4: Simplify the equation
First, calculate the numerator:
\[ 2 \cdot 9.8 \cdot 19.6 = 384.16 \]

Now, divide by g:
\[ R = \frac{384.16}{9.8} \]

### Step 5: Calculate the final result
Now, perform the division:
\[ R = 39.2 \text{ meters} \]

### Conclusion
The projectile will strike the plane after covering a distance of **39.2 meters**.

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