A projectile is fired with a velocity u in such a way that its horizontal range is three times the maximum height aatained . Find the value of horizontal range .

Answer

To solve the problem, we need to use the equations of projectile motion. Let's break down the steps to find the horizontal range when it is three times the maximum height attained.

### Step 1: Understand the relationships
In projectile motion, the horizontal range \( R \) and the maximum height \( H \) are given by the following formulas:
- The horizontal range \( R \) is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
- The maximum height \( H \) is given by:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]

### Step 2: Set up the equation based on the problem statement
According to the problem, the horizontal range is three times the maximum height:
\[ R = 3H \]

### Step 3: Substitute the formulas for \( R \) and \( H \)
Substituting the formulas from Step 1 into the equation from Step 2, we get:
\[ \frac{u^2 \sin 2\theta}{g} = 3 \left(\frac{u^2 \sin^2 \theta}{2g}\right) \]

### Step 4: Simplify the equation
We can simplify this equation by multiplying both sides by \( g \):
\[ u^2 \sin 2\theta = \frac{3u^2 \sin^2 \theta}{2} \]
Now, we can cancel \( u^2 \) from both sides (assuming \( u \neq 0 \)):
\[ \sin 2\theta = \frac{3}{2} \sin^2 \theta \]

### Step 5: Use the double angle identity
Recall that \( \sin 2\theta = 2 \sin \theta \cos \theta \). Substituting this into the equation gives:
\[ 2 \sin \theta \cos \theta = \frac{3}{2} \sin^2 \theta \]

### Step 6: Rearrange the equation
Rearranging the equation, we have:
\[ 4 \sin \theta \cos \theta = 3 \sin^2 \theta \]
Dividing both sides by \( \sin \theta \) (assuming \( \sin \theta \neq 0 \)):
\[ 4 \cos \theta = 3 \sin \theta \]

### Step 7: Solve for \( \tan \theta \)
From the equation \( 4 \cos \theta = 3 \sin \theta \), we can express \( \tan \theta \):
\[ \tan \theta = \frac{4}{3} \]

### Step 8: Find \( R \) in terms of \( u \)
Now, we can find \( R \) using the expression for \( R \):
\[ R = \frac{u^2 \sin 2\theta}{g} \]
Using the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \), we can express \( \sin \theta \) and \( \cos \theta \) in terms of \( \tan \theta \):
\[ \sin \theta = \frac{4}{5}, \quad \cos \theta = \frac{3}{5} \quad (\text{using } \tan \theta = \frac{4}{3}) \]
Thus,
\[ \sin 2\theta = 2 \cdot \frac{4}{5} \cdot \frac{3}{5} = \frac{24}{25} \]

### Step 9: Substitute back to find \( R \)
Now substituting \( \sin 2\theta \) back into the range formula:
\[ R = \frac{u^2 \cdot \frac{24}{25}}{g} = \frac{24u^2}{25g} \]

### Final Answer
Thus, the horizontal range \( R \) is:
\[ R = \frac{24u^2}{25g} \]