the maximum range if a particle projected with a certain speed on a horizontal plane is R.Find its maximum range when projected on an inclined plane with inclination 45°

To determine the maximum range of a particle projected on an inclined plane at a 45° angle, we can start by recalling the formula for the range of a projectile. When a particle is projected on a horizontal plane, its range \( R \) can be expressed as:

Range on a Horizontal Plane

The formula for the range \( R \) of a projectile launched with an initial speed \( v \) at an angle \( \theta \) is given by:

R = \frac{v^2 \sin(2\theta)}{g}

Where \( g \) is the acceleration due to gravity. For a horizontal launch, \( \theta = 0° \), and thus the range is maximized when the angle is 45°.

Understanding the Inclined Plane Scenario

When projecting a particle on an inclined plane, the situation changes slightly. The effective angle of projection and the gravitational component acting along the incline must be considered. For an inclined plane at 45°, we can derive the new range.

Deriving the Range on an Inclined Plane

When a projectile is launched at an angle \( \theta \) with respect to the horizontal, the effective angle of projection relative to the incline also needs to be taken into account. The angle of projection \( \theta \) remains the same, but the incline affects the range calculation.

The range \( R' \) on an inclined plane can be calculated using the formula:

R' = \frac{v^2 \sin(2\theta)}{g \cos(\theta)}

In our case, since the incline is at 45°, we can substitute \( \theta = 45° \):

• For \( \sin(90°) = 1 \)
• For \( \cos(45°) = \frac{1}{\sqrt{2}} \)

Substituting these values into the range formula gives us:

R' = \frac{v^2 \cdot 1}{g \cdot \frac{1}{\sqrt{2}}} = \frac{v^2 \sqrt{2}}{g}

Comparing Ranges

Now, let's compare the ranges:

On a horizontal plane, the maximum range \( R \) is:

R = \frac{v^2}{g}

On the inclined plane, we found:

R' = \frac{v^2 \sqrt{2}}{g}

To find the relationship between \( R' \) and \( R \), we can express \( R' \) in terms of \( R \):

R' = R \cdot \sqrt{2}

Final Thoughts

This means that the maximum range when projected on an inclined plane at 45° is \( \sqrt{2} \) times the maximum range on a horizontal plane. This factor arises from the geometry of the situation and the way gravity acts on the projectile along the incline.

In summary, if a particle has a maximum range \( R \) when projected horizontally, its maximum range when projected on a 45° inclined plane will be \( R' = R \cdot \sqrt{2} \). This relationship highlights the impact of the incline on projectile motion and demonstrates how angles can significantly alter the behavior of projectiles in different contexts.