A particle is projected at an elevation `tan^(-1) ((4)/(3)) from a point O. The ratio of the range on the horizontal plane through O to the greatest height ascended above O is

Options

A. 2
B. 3
C. 4
D. 5

Answer

To solve the problem, we need to find the ratio of the range of a projectile to the maximum height it reaches when projected at an angle of elevation given by \( \theta = \tan^{-1}\left(\frac{4}{3}\right) \).

### Step 1: Identify the angle and calculate its sine and cosine
Given:
\[ \tan \theta = \frac{4}{3} \]
From this, we can construct a right triangle where the opposite side is 4 and the adjacent side is 3. Using the Pythagorean theorem, we can find the hypotenuse:
\[ \text{Hypotenuse} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

Now, we can find \( \sin \theta \) and \( \cos \theta \):
\[ \sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{4}{5} \]
\[ \cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{3}{5} \]

### Step 2: Calculate the range (R)
The formula for the range \( R \) of a projectile is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
Using the double angle identity, \( \sin 2\theta = 2 \sin \theta \cos \theta \):
\[ \sin 2\theta = 2 \cdot \frac{4}{5} \cdot \frac{3}{5} = \frac{24}{25} \]
Thus, the range becomes:
\[ R = \frac{u^2 \cdot \frac{24}{25}}{g} \]

### Step 3: Calculate the maximum height (H)
The formula for the maximum height \( H \) is given by:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
Calculating \( \sin^2 \theta \):
\[ \sin^2 \theta = \left(\frac{4}{5}\right)^2 = \frac{16}{25} \]
Thus, the maximum height becomes:
\[ H = \frac{u^2 \cdot \frac{16}{25}}{2g} = \frac{u^2 \cdot \frac{16}{25}}{2g} = \frac{8u^2}{25g} \]

### Step 4: Find the ratio of range to maximum height
Now, we find the ratio \( \frac{R}{H} \):
\[ \frac{R}{H} = \frac{\frac{u^2 \cdot \frac{24}{25}}{g}}{\frac{8u^2}{25g}} = \frac{u^2 \cdot \frac{24}{25}}{g} \cdot \frac{25g}{8u^2} \]
The \( u^2 \) and \( g \) cancel out:
\[ \frac{R}{H} = \frac{24}{8} = 3 \]

### Final Answer
The ratio of the range on the horizontal plane through O to the greatest height ascended above O is:
\[ \boxed{3} \]