Physics > Motion in Two Dimension > 5.0 Projectile on an inclined plane
Motion in Two Dimension
1.0 Introduction
2.0 Projectile motion
3.0 Ground to ground projectile motion
3.1 Maximum height
3.2 Time of flight
3.3 Range
3.4 Trajectory of a projectile
3.5 Summary
3.6 Solved Examples
4.0 Projectile thrown parallel to the horizontal
5.0 Projectile on an inclined plane
6.0 Relative motion between two projectiles
5.3 Solved examples
3.2 Time of flight
3.3 Range
3.4 Trajectory of a projectile
3.5 Summary
3.6 Solved Examples
Question: A ball is projected from point $A$ with a velocity of $10 \ m/s$ perpendicular to the inclined plane as shown in the figure. Find the range and time of flight of the ball on the inclined plane.
Solution:
| Horizontal motion | Vertical motion |
| ${u_X} = 0$ | ${u_Y} = u$ |
| ${a_X} = + g\sin \alpha $ | ${a_Y} = - g\cos \alpha $ |
Time of flight
The time in which the particle returns back on the inclined plane is the time of flight.
${\overrightarrow s _Y} = 0$
${\overrightarrow u _Y} = +u=10$
${\overrightarrow a _Y} = - g \cos \alpha$
${\overrightarrow u _Y} = +u=10$
${\overrightarrow a _Y} = - g \cos \alpha$
$t=?$
We can write, $${\overrightarrow s _Y} = {\overrightarrow u _Y}t + \frac{1}{2}{\overrightarrow a _Y}{t^2}$$$$0 = ut + \frac{1}{2}\left( { - g\cos \alpha } \right){t^2}$$$$t = \frac{{2u}}{{g\cos \alpha }}$$$$t = \frac{{2 \times 10}}{{10 \times \cos 30^\circ }}$$$$t = \frac{4}{{\sqrt 3 }}s$$
Range
${\overrightarrow s _X} = +R=?$
${\overrightarrow u _X} = 0$
${\overrightarrow a _X} = = g \sin \alpha$
${\overrightarrow u _X} = 0$
${\overrightarrow a _X} = = g \sin \alpha$
$t =T= \frac{4}{{\sqrt 3 }}s$
We can write,
$${\overrightarrow s _X} = {\overrightarrow u _X}t + \frac{1}{2}{\overrightarrow a _X}{t^2}$$$$ + R = 0 + \frac{1}{2}\left( { + g\sin \alpha } \right){\left( {\frac{4}{{\sqrt 3 }}} \right)^2}$$$$R = \frac{{10 \times \frac{1}{2}}}{2} \times \frac{{16}}{3}$$$$R = \frac{{40}}{3}m$$
Question: A particle is projected with velocity $u$ at an angle $\theta$ from the inclined plane of inclination $\alpha$. The particle hits the plane perpendicularly. Find the time of flight.
Solution:
$X$ Horizontal motion | $Y$ Vertical motion |
| ${\overrightarrow u _X} = + u \cos \theta$ | ${\overrightarrow u _Y} = +u \sin \theta$ |
| ${\overrightarrow a _X} = -g \sin \alpha$ | ${\overrightarrow a _Y} = - g \cos \alpha$ |
When the particle hits the inclined plane perpendicularly, its velocity along the inclined plane is zero.
Mathematically,
$${\overrightarrow v _X} = 0$$
So, we can write,
$${\overrightarrow v _X} = {\overrightarrow u _X} + {\overrightarrow a _X}t$$$$0 = u\cos \theta + \left( { - g\sin \alpha } \right)t$$$$t = \frac{{u\cos \theta }}{{g\sin \alpha }}$$
