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

Explanation

To solve the problem of how far the projectile will travel before striking the ground, we can follow these steps:

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

### Step 2: Calculate the time of flight
The time of flight (T) for a projectile can be calculated using the formula:
\[ T = \frac{2 \cdot u_y}{g} \]
Substituting the values:
\[ T = \frac{2 \cdot 19.6}{9.8} = \frac{39.2}{9.8} = 4 \text{ seconds} \]

### Step 3: Calculate the horizontal distance (range)
The horizontal distance (R) covered by the projectile can be calculated using the formula:
\[ R = u_x \cdot T \]
Substituting the values:
\[ R = 9.8 \cdot 4 = 39.2 \text{ meters} \]

### Step 4: Conclusion
The projectile will strike the plane after covering a distance of **39.2 meters**.

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