A particle projected with velocity 20 m/s at an angle 53(degree) with the horizontal strikes an inclined plane of inclination 60 ( degree) perpendicularly . find its speed at the point of striking.

To determine the speed of a particle projected at a velocity of 20 m/s at an angle of 53 degrees with the horizontal when it strikes an inclined plane of 60 degrees perpendicularly, we can break this problem down into several steps. We will analyze the motion of the particle and apply some principles of physics, particularly those related to projectile motion and vector components.

Breaking Down the Problem

First, let's identify the components of the initial velocity. The particle is projected at an angle, so we can resolve the initial velocity into horizontal and vertical components using trigonometric functions.

Calculating Velocity Components

The initial velocity \( v_0 \) is 20 m/s. We can find the horizontal (\( v_{0x} \)) and vertical (\( v_{0y} \)) components as follows:

• Horizontal Component: \( v_{0x} = v_0 \cdot \cos(\theta) = 20 \cdot \cos(53^\circ) \)
• Vertical Component: \( v_{0y} = v_0 \cdot \sin(\theta) = 20 \cdot \sin(53^\circ) \)

Using the cosine and sine values:

• \( \cos(53^\circ) \approx 0.6018 \)
• \( \sin(53^\circ) \approx 0.7986 \)

Calculating these components:

• \( v_{0x} \approx 20 \cdot 0.6018 \approx 12.036 \, \text{m/s} \)
• \( v_{0y} \approx 20 \cdot 0.7986 \approx 15.972 \, \text{m/s} \)

Time of Flight Until Impact

Next, we need to find the time it takes for the particle to hit the inclined plane. The inclined plane makes an angle of 60 degrees with the horizontal. The condition for the particle to strike the plane perpendicularly means that the angle of the velocity vector at the point of impact must equal the angle of the incline.

The slope of the inclined plane can be expressed as the tangent of the angle of inclination:

\( \tan(60^\circ) = \sqrt{3} \approx 1.732 \)

At the point of impact, the vertical and horizontal components of the velocity must satisfy the relationship:

\( \frac{v_{y}}{v_{x}} = \tan(60^\circ) \)

Thus, we can express the vertical component of the velocity at impact as:

\( v_{y} = \sqrt{3} \cdot v_{x} \)

Using Kinematic Equations

To find the vertical component of the velocity just before impact, we can use the kinematic equation for vertical motion:

\( v_{y} = v_{0y} - g t \)

Where \( g \) is the acceleration due to gravity (approximately 9.81 m/s²). We can also express the horizontal motion as:

\( v_{x} = v_{0x} \) (since there is no horizontal acceleration)

Setting Up the Equations

Now we can substitute \( v_{y} \) in terms of \( v_{x} \) into the vertical motion equation:

\( \sqrt{3} \cdot v_{x} = v_{0y} - g t \)

Substituting \( v_{x} \) with \( v_{0x} \):

\( \sqrt{3} \cdot v_{0x} = v_{0y} - g t \)

Now we can solve for \( t \) using the horizontal motion:

\( x = v_{0x} t \) (where \( x \) is the horizontal distance traveled)

We can express \( t \) as:

\( t = \frac{x}{v_{0x}} \)

Finding the Speed at Impact

Finally, we can find the speed at the point of impact by using the Pythagorean theorem to combine the horizontal and vertical components of the velocity:

\( v = \sqrt{v_{x}^2 + v_{y}^2} \)

Substituting the values we have calculated, we can find the speed at the point of striking the inclined plane. Since we know \( v_{x} \) and can find \( v_{y} \) using the relationship established earlier, we can compute the final speed.

In summary, the speed of the particle at the point of striking the inclined plane can be calculated using the components of the velocity just before impact, ensuring that the angle conditions are satisfied. This approach combines kinematics with trigonometry to yield the desired result.