What do you mean by projectile motion and projectile particle? Find the value of the position of a projectile particle at any instant of time.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) Projectile Motion: When an object is thrown into the Earth's gravitational field,it moves with a constant horizontal velocity and a constant vertical acceleration. Such a two-dimensional motion is called projectile motion,and the object is called a projectile.
For example,when we kick a football,it performs projectile motion if air resistance is neglected.
$A$ projectile has motion along a horizontal path with uniform velocity $(a_x = 0)$.
$A$ projectile has motion along a vertical path under gravity with uniform acceleration equal to $g$ $(a_y = -g)$.
Suppose the projectile is launched with velocity $\vec{v}_0$ at an angle $\theta_0$ with the $X$-axis. The acceleration is $\vec{a} = -g \hat{j}$.
The initial velocity components are:
$v_{0x} = v_0 \cos \theta_0$
$v_{0y} = v_0 \sin \theta_0$
Using the kinematic equation $r = r_0 + v_0 t + \frac{1}{2} a t^2$:
For the $X$-coordinate:
$x(t) = x_0 + v_{0x} t + \frac{1}{2} a_x t^2 = 0 + (v_0 \cos \theta_0) t + 0 = v_0 \cos \theta_0 t$
For the $Y$-coordinate:
$y(t) = y_0 + v_{0y} t + \frac{1}{2} a_y t^2 = 0 + (v_0 \sin \theta_0) t - \frac{1}{2} g t^2 = v_0 \sin \theta_0 t - \frac{1}{2} g t^2$
Thus,the position of the particle at any instant $t$ is given by the coordinates $(x(t), y(t)) = (v_0 \cos \theta_0 t, v_0 \sin \theta_0 t - \frac{1}{2} g t^2)$.

Explore More

Similar Questions

The path of a projectile is given by the equation $y = ax - bx^2$, where $a$ and $b$ are constants, and $x$ and $y$ are the horizontal and vertical distances of the projectile from the point of projection, respectively. The maximum height attained by the projectile and the angle of projection are respectively:

$A$ stone projected from the ground with a velocity $50 \ m/s$ at an angle of $30^{\circ}$ with the horizontal crosses a wall after a time of $3 \ s$. Then the horizontal distance beyond the wall that the stone strikes the ground is (acceleration due to gravity $g = 10 \ m/s^2$) (in $m$)

The equation of a projectile is given by $y = Px - Qx^2$,where $P$ and $Q$ are constants. The ratio of the maximum height to the range of the projectile is

$A$ projectile object is thrown in the upward direction making an angle of $60^{\circ}$ with the horizontal with velocity of $140 \ m/s$. Then, the time after which its velocity makes an angle $30^{\circ}$ with the horizontal is (Take, $g=10 \ m/s^2$)

If the initial kinetic energy of a projectile is $100 \, J$ and the kinetic energy at the maximum height is $30 \, J$,what is the angle of projection?

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo