Find time of flight if projectile thrown horizontally with speed `10 ms^(-1)` from a long inclined plane which makes an angle of `theta = 45^(@)` from horizontal.

Options

A. `sqrt(2)sec`
B. `2sqrt(2)sec`
C. `2sec`
D. none

Explanation

To find the time of flight of a projectile thrown horizontally from a long inclined plane at an angle of \( \theta = 45^\circ \) with a speed of \( 10 \, \text{m/s} \), we can follow these steps:

### Step 1: Understand the Problem
We have a projectile launched horizontally from an inclined plane. The angle of the incline is \( 45^\circ \). The initial speed of the projectile is \( 10 \, \text{m/s} \).

### Step 2: Resolve the Initial Velocity
Since the projectile is thrown horizontally, its initial vertical component of velocity (\( u_y \)) is \( 0 \, \text{m/s} \). The horizontal component of velocity (\( u_x \)) is \( 10 \, \text{m/s} \).

### Step 3: Determine the Acceleration Components
The acceleration due to gravity (\( g \)) acts downward. We need to resolve this into components along the incline. The component of gravitational acceleration acting perpendicular to the incline is:
\[ g_{\perpendicular} = g \cos(45^\circ) = \frac{g}{\sqrt{2}} \]
And the component acting parallel to the incline is:
\[ g_{\parallel} = g \sin(45^\circ) = \frac{g}{\sqrt{2}} \]

### Step 4: Calculate Time of Flight
Since the projectile is thrown horizontally, we need to find the time it takes to fall vertically until it reaches the incline. The vertical motion can be described by:
\[ s = u_y t + \frac{1}{2} g_{\perpendicular} t^2 \]
Here, \( u_y = 0 \), so the equation simplifies to:
\[ s = \frac{1}{2} g_{\perpendicular} t^2 \]
The distance \( s \) that the projectile falls vertically until it hits the incline can be expressed in terms of the angle of the incline. For a \( 45^\circ \) incline, the vertical distance fallen is equal to the horizontal distance traveled.

### Step 5: Set Up the Relationship
Since the incline is at \( 45^\circ \), the horizontal distance traveled \( x \) is equal to the vertical distance fallen \( y \):
\[ x = y \]
Using the relationship from the projectile motion:
\[ x = u_x t \]
And for the vertical distance:
\[ y = \frac{1}{2} g_{\perpendicular} t^2 \]
Setting \( x = y \):
\[ u_x t = \frac{1}{2} g_{\perpendicular} t^2 \]

### Step 6: Solve for Time of Flight
Substituting \( u_x = 10 \, \text{m/s} \) and \( g_{\perpendicular} = \frac{g}{\sqrt{2}} \):
\[ 10 t = \frac{1}{2} \left(\frac{g}{\sqrt{2}}\right) t^2 \]
\[ 10 t = \frac{g t^2}{2\sqrt{2}} \]
Rearranging gives:
\[ t = \frac{20\sqrt{2}}{g} \]
Substituting \( g \approx 9.81 \, \text{m/s}^2 \):
\[ t \approx \frac{20\sqrt{2}}{9.81} \approx 2.87 \, \text{s} \]

### Final Answer
The time of flight is approximately \( 2.87 \, \text{s} \).

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