Name: Motion on an Inclined Plane
Uploaded: 2019-07-19
Duration: 6 min 34 s
Description: We all are curious about gravity. Through this video, let us learn to break the force of gravity into components in case of motion on an inclined plane.

Question

If

h

be the maximum height of a projectile moving under the gravitational field of earth, then prove that its velocity of projection will be

\frac{\sqrt{2 g h}}{\sin θ}

, where

θ

is the angle of projection.

Accepted Answer

When a body is thrown upward in a direction different from the vertical, then it is acted upon by acceleration due to gravity (which acts vertically downwards) and moves along a curved path in a vertical plane. This motion is called the 'projectile motion' and its path is called the 'trajectory'.

Suppose a body is projected up with an initial velocity $u$ in a direction making an angle $θ_{0}$ with the horizontal. Resolving $u$ in horizontal and vertical components, we have,
$u_{x} = ucosθ$ and

$u_{y} = usinθ$
The body moves under constant gravitational acceleration $g$ acting vertically downwards. Therefore, the horizontal velocity remains unchanged throughout the motion (provided the resistance of air is negligible). But the vertical velocity continuously changes.

Let $h$ be the vertical height of the highest point $P$ . The vertical velocity at $P$ is zero. Hence, in equation $v^{2} = u^{2} - 2 gh$ , substituting $v = v_{y} = 0$ and $u = u_{y} = {sinθ}_{0}$ , we get,
$0 = ({usinθ}_{0})^{2} - 2 gh$

or $u^{2} \sin^{2} θ_{0} = 2 gh$

or $u = \frac{\sqrt{2 gh}}{{sinθ}_{0}}$

If $h$ be the maximum height of a projectile moving under the gravitational field of earth, then prove that its velocity of projection will be
$\frac{\sqrt{2 g h}}{\sin θ}$ , where $θ$ is the angle of projection.

Important Questions on Motion in One and Two Dimensions

Learn and Prepare for any exam you want !

If h be the maximum height of a projectile moving under the gravitational field of earth, then prove that its velocity of projection will be 2ghsinθ, where θ is the angle of projection.

Important Questions on Motion in One and Two Dimensions

Learn and Prepare for any exam you want !

If $h$ be the maximum height of a projectile moving under the gravitational field of earth, then prove that its velocity of projection will be
$\frac{\sqrt{2 g h}}{\sin θ}$ , where $θ$ is the angle of projection.