The speed at the maximum height of a projectile is `(1)/(3)` of its initial speed u. If its range on the horizontal plane is `(n sqrt(2)u^(2))/(9g)` then value of n is

Explanation

To solve the problem step by step, let's break it down:

### Step 1: Understand the given information
We know that at maximum height, the speed of the projectile is \( \frac{1}{3} \) of its initial speed \( u \). This means:
\[ v = \frac{1}{3} u \]
At maximum height, the vertical component of the velocity becomes zero, and the horizontal component remains \( u \cos \theta \). Therefore, we have:
\[ u \cos \theta = \frac{1}{3} u \]

### Step 2: Simplify the equation
Dividing both sides by \( u \) (assuming \( u \neq 0 \)):
\[ \cos \theta = \frac{1}{3} \]

### Step 3: Use the identity for sine
Using the Pythagorean identity, we can find \( \sin \theta \):
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
Substituting \( \cos \theta \):
\[ \sin^2 \theta + \left(\frac{1}{3}\right)^2 = 1 \]
\[ \sin^2 \theta + \frac{1}{9} = 1 \]
\[ \sin^2 \theta = 1 - \frac{1}{9} = \frac{8}{9} \]
\[ \sin \theta = \frac{2\sqrt{2}}{3} \]

### Step 4: Calculate \( \sin 2\theta \)
Using the double angle formula:
\[ \sin 2\theta = 2 \sin \theta \cos \theta \]
Substituting the values we found:
\[ \sin 2\theta = 2 \left(\frac{2\sqrt{2}}{3}\right) \left(\frac{1}{3}\right) = \frac{4\sqrt{2}}{9} \]

### Step 5: Calculate the range \( R \)
The range \( R \) of the projectile is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
Substituting \( \sin 2\theta \):
\[ R = \frac{u^2 \cdot \frac{4\sqrt{2}}{9}}{g} = \frac{4\sqrt{2} u^2}{9g} \]

### Step 6: Compare with the given range
We are given that the range can also be expressed as:
\[ R = \frac{n \sqrt{2} u^2}{9g} \]
Setting the two expressions for \( R \) equal to each other:
\[ \frac{4\sqrt{2} u^2}{9g} = \frac{n \sqrt{2} u^2}{9g} \]

### Step 7: Solve for \( n \)
Canceling \( \sqrt{2} u^2 \) and \( 9g \) from both sides gives:
\[ 4 = n \]

### Final Answer
Thus, the value of \( n \) is:
\[ \boxed{4} \]