A projectile is projected with speed u at an angle of `60^@` with horizontal from the foot of an inclined plane. If the projectile hits the inclined plane horizontally, the range on inclined plane will be.

Options

A. `(u^2(sqrt21))/(2g)`
B. `(3u^2)/(4g)`
C. `u^2/(2beta)`
D. `((sqrt21)u^2)/(8g)`

Explanation

To solve the problem of finding the range of a projectile projected at an angle of \(60^\circ\) to the horizontal from the foot of an inclined plane, we can follow these steps:

### Step 1: Resolve the Initial Velocity
The initial velocity \(u\) can be resolved into horizontal and vertical components:
- Horizontal component:
\[ u_x = u \cos 60^\circ = \frac{u}{2} \]
- Vertical component:
\[ u_y = u \sin 60^\circ = \frac{\sqrt{3}u}{2} \]

### Step 2: Understand the Projectile Motion
When the projectile hits the inclined plane horizontally, it means that at the point of impact, the vertical component of the velocity \(V_y = 0\). This occurs at the maximum height of the projectile.

### Step 3: Calculate the Maximum Height
The maximum height \(H\) reached by the projectile can be calculated using the formula:
\[ H = \frac{u_y^2}{2g} = \frac{\left(\frac{\sqrt{3}u}{2}\right)^2}{2g} = \frac{3u^2}{8g} \]

### Step 4: Calculate the Horizontal Range on the Inclined Plane
The horizontal range \(R\) on the inclined plane can be calculated using the formula for the range of a projectile:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
For our case, \(\theta = 60^\circ\):
\[ R = \frac{u^2 \sin 120^\circ}{g} = \frac{u^2 \cdot \frac{\sqrt{3}}{2}}{g} = \frac{\sqrt{3}u^2}{2g} \]
Since we are interested in the range on the inclined plane, we need to find half of this range:
\[ \frac{R}{2} = \frac{\sqrt{3}u^2}{4g} \]

### Step 5: Use Pythagorean Theorem
Let’s denote the height \(H\) and the horizontal distance \(D\) as:
- \(AB = H = \frac{3u^2}{8g}\)
- \(BC = \frac{R}{2} = \frac{\sqrt{3}u^2}{4g}\)

Now, we can find the length of the incline \(AC\) using the Pythagorean theorem:
\[ AC = \sqrt{AB^2 + BC^2} \]
Substituting the values:
\[ AC = \sqrt{\left(\frac{3u^2}{8g}\right)^2 + \left(\frac{\sqrt{3}u^2}{4g}\right)^2} \]
Calculating:
\[ = \sqrt{\frac{9u^4}{64g^2} + \frac{3u^4}{16g^2}} = \sqrt{\frac{9u^4}{64g^2} + \frac{12u^4}{64g^2}} = \sqrt{\frac{21u^4}{64g^2}} = \frac{u^2\sqrt{21}}{8g} \]

### Final Answer
Thus, the range on the inclined plane is:
\[ \text{Range on inclined plane} = \frac{u^2\sqrt{21}}{8g} \]

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