A ball is launched along a smooth inclined plane at a speed of v. How much horizontal distance will it travel before it rolls off the plane? The plane is inclined to the horizon at an angle of 45^\circ . The initial velocity of the ball forms an angle of 45^\circ with the horizontal edge of the plane.

Solutions of Savchenko Problems in Physics

Aliaksandr Melnichenka
October 2023

    <h3 id="back-link"><a href="/#1.3">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $1.3.7.$ A ball is launched along a smooth inclined plane at a speed of $v$. How much horizontal distance will it travel before it rolls off the plane? The plane is inclined to the horizon at an angle of $45^\circ$ . The initial velocity of the ball forms an angle of $45^\circ$ with the horizontal edge of the plane.

    <h3>Solution</h3>
    <p>

On an inclined plane, the body is acted upon by the component of acceleration due to gravity: $$a = g^* = g \cdot \cos{45^{\circ}}$$ From the formula for flight range obtained in 1.3.6 it follows: $$L=\frac{v^2}{g^*}\operatorname{sin}90^{\circ}$$ We substitute $g^*$: $$\fbox{$L=\frac{v^2 \sqrt{2}}{g}$}$$

    </p>

    <h4>Answer</h4>
    <p>
        $$L= \sqrt {2} v^{2} / g$$
    </p>


<footer class="row container">
  <br>
    <p>
        <small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small>
    </p>
    <p>
        <small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
    </p>
</footer>