Askiitians Tutor Team
Last Activity: 4 Months ago
To solve this problem, we need to analyze the motion of the projectile and its interaction with the inclined plane. We can break this down into manageable steps, focusing on the components of motion and the geometry involved.
Understanding the Motion of the Projectile
The particle is projected with an initial velocity \( u = 50 \, \text{m/s} \) at an angle of \( 53^\circ \) to the horizontal. The incline makes an angle of \( 30^\circ \) with the horizontal. We can start by determining the equations of motion for the projectile.
Breaking Down the Initial Velocity
We can resolve the initial velocity into its horizontal and vertical components:
- Horizontal component: \( u_x = u \cdot \cos(53^\circ) \)
- Vertical component: \( u_y = u \cdot \sin(53^\circ) \)
Calculating these values:
- \( u_x = 50 \cdot \cos(53^\circ) \approx 30 \, \text{m/s} \)
- \( u_y = 50 \cdot \sin(53^\circ) \approx 40 \, \text{m/s} \)
Finding the Equation of the Incline
The equation of the incline can be expressed in terms of \( y \) (vertical position) and \( x \) (horizontal position). The slope of the incline is given by the angle \( 30^\circ \), so the equation can be written as:
\( y = \tan(30^\circ) \cdot x \)
Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), we have:
\( y = \frac{1}{\sqrt{3}} x \)
Projectile Motion Equations
The vertical position of the projectile as a function of time \( t \) is given by:
\( y(t) = u_y \cdot t - \frac{1}{2} g t^2 \)
Substituting the values, we have:
\( y(t) = 40t - 4.9t^2 \)
Setting the Equations Equal
To find the time when the projectile hits the incline, we set the equations for \( y \) equal to each other:
\( 40t - 4.9t^2 = \frac{1}{\sqrt{3}} (30t) \)
Rearranging gives:
\( 40t - 4.9t^2 - \frac{30}{\sqrt{3}} t = 0 \)
Factoring out \( t \):
\( t(40 - 4.9t - \frac{30}{\sqrt{3}}) = 0 \)
We discard \( t = 0 \) (the initial time) and solve the quadratic equation:
\( 4.9t^2 - (40 - \frac{30}{\sqrt{3}})t = 0 \)
Calculating Time of Flight
Using the quadratic formula, we can find the positive root for \( t \). After solving, we find that:
\( t \approx 8 - 2\sqrt{3} \, \text{s} \)
Determining the Angle Beta
Next, we need to find the angle \( \beta \) that the projectile makes with the normal to the incline at the point of impact. The normal to the incline makes an angle of \( 30^\circ \) with the horizontal.
The slope of the projectile's trajectory at impact can be found using the derivative of its vertical position:
\( \text{slope} = \frac{dy}{dx} = \frac{u_y - gt}{u_x} \)
Substituting the values at \( t = 8 - 2\sqrt{3} \):
\( \text{slope} = \frac{40 - 9.8(8 - 2\sqrt{3})}{30} \)
Calculating this gives us the slope of the projectile at impact. The angle \( \theta \) that the trajectory makes with the horizontal can be found using:
\( \tan(\theta) = \text{slope} \)
Finally, the angle \( \beta \) is given by:
\( \beta = \theta - 30^\circ \)
Final Thoughts
By following these steps, we can determine both the time taken for the projectile to hit the incline and the angle \( \beta \) it makes with the normal. This approach combines kinematic equations with geometric relationships, providing a comprehensive understanding of projectile motion in relation to inclined surfaces.
