A projectille covers a horizontal distance of 6 m, when it reaches 10 m above the level of projection, how farther should if travel horizontally to reach back the same height , if its horizontal range is 20 m ?

Options

A. 6 m
B. 12 m
C. 8 m
D. 20 m

Answer

To solve the problem step by step, we can follow these steps:

### Step 1: Understand the Problem
We have a projectile that covers a horizontal distance of 6 meters when it reaches a height of 10 meters. The total horizontal range of the projectile is 20 meters. We need to find out how much farther it should travel horizontally to return to the same height of 10 meters.

### Step 2: Identify the Points
Let’s label the points:
- Point A: the point of projection (ground level)
- Point B: the point where the projectile reaches a height of 10 meters and is 6 meters horizontally from A
- Point C: the point where the projectile reaches the maximum height (10 meters) again before coming down
- Point D: the point where the projectile lands back on the ground

### Step 3: Analyze the Horizontal Motion
The horizontal motion of the projectile is uniform because there is no horizontal acceleration. The horizontal distance covered when the projectile reaches point B is 6 meters.

### Step 4: Use the Total Range
The total horizontal range of the projectile is given as 20 meters. This means:
- The total horizontal distance from point A to point D is 20 meters.

### Step 5: Calculate the Remaining Distance
From point A to point B, the projectile has traveled 6 meters. The distance from point B to point D is also 6 meters (as it will take the same horizontal distance to come down to the same height).

So, we can set up the equation:
\[ \text{Distance from A to B} + \text{Distance from B to C} + \text{Distance from C to D} = \text{Total Range} \]
\[ 6 \, \text{m} + \text{CD} + 6 \, \text{m} = 20 \, \text{m} \]

### Step 6: Solve for CD
Now, substituting the known values:
\[ 6 + \text{CD} + 6 = 20 \]
\[ \text{CD} + 12 = 20 \]
\[ \text{CD} = 20 - 12 \]
\[ \text{CD} = 8 \, \text{m} \]

### Conclusion
The projectile should travel an additional horizontal distance of **8 meters** to reach back to the same height of 10 meters.

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