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.
When an object is thrown in gravitational field of the Earth, it moves with constant horizontal velocity and constant vertical acceleration. Such a two dimensional motion is called a projectile motion and such an 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 horizontal path with uniform velocity.
A projectile has motion along vertical path under gravity with uniform acceleration equal to ' $g$ '.
Suppose that the projectile is launched with velocity $\vec{v}_{0}$ that makes an angle $\theta_{0}$ with the $\mathrm{X}$-axis as show in figure.
$\mathrm{g}$ acts on the body along ' $\mathrm{Y}$ ' axis and in opposite direction
The acceleration ' $a$ ' is given by
$\therefore \vec{a}=-\mathrm{g} \hat{j}$
here $a_{x}=0, a_{y}=-g$
The velocity $\vec{v}_{0}$ can be divided into two components,
$v_{\mathrm{o} x}=v_{\mathrm{o}} \cos \theta_{\mathrm{o}}$
$v_{\mathrm{oy}}=v_{\mathrm{o}} \sin \theta_{\mathrm{o}}$
If we take the initial position to be the origin of the coordinate system, the coordinates of the point of projection would be :
$x_{0}=0$, and $y_{0}=0$
Two particles start simultaneously from the same point and move along two straight lines, one with uniform velocity $v$ and other with a uniform acceleration $a.$ If $\alpha$ is the angle between the lines of motion of two particles then the least value of relative velocity will be at time given by
Velocity of a particle moving in a curvilinear path in a horizontal $X$ $Y$ plane varies with time as $\vec v = (2t\hat i + t^2 \hat j) \ \ m/s.$ Here, $t$ is in second. At $t = 1\ s$
A particle is moving with velocity $\vec v = K(y\hat i + x\hat j)$ where $K$ is a constant. The general equation for its path is
A particle moves from the point $\left( {2.0\hat i + 4.0\hat j} \right)\,m$, at $t = 0$ with an initial velocity $\left( {5.0\hat i + 4.0\hat j} \right)\,m{s^{ - 1 }}$. It is acted upon by a constant force which produces a constant acceleration $\left( {4.0\hat i + 4.0\hat j} \right)\,m{s^{ - 2}}$. What is the distance of the particle from the origin at time $2\,s$
A particle starts from origin at $t=0$ with a velocity $5.0 \hat{ i }\; m / s$ and moves in $x-y$ plane under action of a force which produces a constant acceleration of $(3.0 \hat{ i }+2.0 \hat{ j })\; m / s ^{2} .$
$(a)$ What is the $y$ -coordinate of the particle at the instant its $x$ -coordinate is $84 \;m$ ?
$(b)$ What is the speed of the particle at this time?