An object is projected at an angle of `45^(@)` with the horizontal. The horizontal range and the maximum height reached will be in the ratio.

Options

A. `1:2`
B. `2:1`
C. `1:4`
D. `4:1`

Answer

To find the ratio of the horizontal range (R) to the maximum height (H) for an object projected at an angle of 45 degrees, we can use the following steps:

### Step 1: Understand the basic equations of projectile motion
In projectile motion, the horizontal range (R) and the maximum height (H) can be calculated using the initial velocity (u) and the angle of projection (θ).

### Step 2: Calculate the horizontal range (R)
The formula for the horizontal range (R) of a projectile is given by:

\[ R = \frac{u^2 \sin(2\theta)}{g} \]

For an angle of θ = 45 degrees, we have:

\[ \sin(2 \times 45^\circ) = \sin(90^\circ) = 1 \]

Thus, the formula for R simplifies to:

\[ R = \frac{u^2}{g} \]

### Step 3: Calculate the maximum height (H)
The formula for the maximum height (H) of a projectile is given by:

\[ H = \frac{u^2 \sin^2(\theta)}{2g} \]

Again, for θ = 45 degrees:

\[ \sin(45^\circ) = \frac{1}{\sqrt{2}} \]

So,

\[ H = \frac{u^2 \left(\frac{1}{\sqrt{2}}\right)^2}{2g} = \frac{u^2 \cdot \frac{1}{2}}{2g} = \frac{u^2}{4g} \]

### Step 4: Find the ratio R:H
Now, we can find the ratio of the horizontal range (R) to the maximum height (H):

\[ \text{Ratio} = \frac{R}{H} = \frac{\frac{u^2}{g}}{\frac{u^2}{4g}} = \frac{u^2}{g} \cdot \frac{4g}{u^2} = 4 \]

### Conclusion
The ratio of the horizontal range (R) to the maximum height (H) for an object projected at an angle of 45 degrees is:

\[ R : H = 4 : 1 \]