Equation Of Path Of A Projectile

Equation of Path of a Projectile

$
y=x \tan \theta-\frac{g x^2}{2 u^2 \cos ^2 \theta}
$$
or-
$$
y=x \tan \theta\left(1-\frac{x}{R}\right)
$

Where R is the horizontal range of the projectile.

It is the equation of the parabola, So the trajectory/path of the projectile is

parabolic in nature

$g \rightarrow$ Acceleration due to gravity
$u \rightarrow$ initial velocity
$\theta=$ Angle of projection

Solved Question Besed On Equation Of Path Of A Projectile

Example 1: The trajectory of a projectile near the surface of the earth is given as $y=2 x-9 x^2$. If it were launched at an angle $\theta_0$ with speed $v_0$ then $\left(g=10 \mathrm{~ms}^{-2}\right)$ :

1) $\theta_0=\sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)$ and $v_0=\frac{3}{5} m s^{-1}$
2) $\theta_0=\cos ^{-1}\left(\frac{2}{\sqrt{5}}\right)$ and $v_0=\frac{3}{5} m s^{-1}$
3) $\theta_0=\cos ^{-1}\left(\frac{1}{\sqrt{5}}\right)$ and $v_0=\frac{5}{3} m s^{-1}$
4) $\theta_0=\sin ^{-1}\left(\frac{1}{\sqrt{5}}\right)$ and $v_0=\frac{5}{3} m s^{-1}$

Solution:

Given :
$
y=2 x-9 x^2
$

The standard equation of trajectory for a projectile
$
y=x \tan \theta_0-\frac{g x^2}{2 v_0^2 \cos ^2 \theta_0}
$

On comparing Eq. (1) and (2)
$
\begin{aligned}
& \tan \theta_0=2 \\
& \Rightarrow \theta_0=\sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)=\cos ^{-1}\left(\frac{1}{\sqrt{5}}\right) \\
& \frac{g}{2 v_0^2 \cos ^2 \theta_0}=9 \\
& \Rightarrow \frac{g}{2 v_0^2 \times \frac{1}{5}}=9 \\
& \Rightarrow \frac{10 \times 5}{2 \times 9}=v_0^2 \\
& \Rightarrow v_0^2=\frac{25}{9} \\
& v_0=\frac{5}{3} \mathrm{~m} / \mathrm{s}
\end{aligned}
$

Hence, the answer is the Option (3).

Example 2: A particle just clears a wall of height b at a distance a and strikes the ground at a distance c from the point of projection. The angle of projection is:

1) $\tan ^{-1}\left(\frac{b c}{a(c-a)}\right)$
2) $\tan ^{-1}\left(\frac{b c}{a}\right)$
3) $\tan ^{-1}\left(\frac{b}{a * c}\right)$
4) $45^{\circ}$

Solution:

$\begin{aligned}
& a=(u \cos \alpha) t \text { and } \\
& b=(u \sin \alpha) t-\frac{1}{2} g t^2 \\
& b=a \tan \alpha-\frac{1}{2} g \frac{a^2}{u^2 \cos ^2 \alpha} \\
& \text { also, } c=\frac{u^2 \sin 2 \alpha}{g} \\
& \Rightarrow b=a \tan \alpha-\frac{a^2 g}{2}\left(\frac{\sin 2 \alpha}{c g}\right) \sec ^2 \alpha \\
& \Rightarrow b=a \tan \alpha-\frac{a^2}{2 c} 2 \tan \alpha \\
& \Rightarrow\left(a-\frac{a^2}{c}\right) \tan \alpha=b \\
& \Rightarrow \tan \alpha=\frac{b c}{a(c-a)}
\end{aligned}$

Hence, the answer is option (1).

Example 3: $ \text { Position of a particle moving in } x-y \text { plane as a function of time } t \text { is } 2 t \hat{i}+4 t^2 \hat{j} \text {. The equation of the trajectory of the particle is : }$

1) $ y=x^2 $
2) $ y=2 x $
3) $ y^2=x $
4) $ y=x $

Solution:

$
\begin{aligned}
& \mathrm{x}=2 \mathrm{t} \Rightarrow \mathrm{t}=\mathrm{x} 2 \\
& \mathrm{y}=4 \mathrm{t}^2 \\
& \text { put } \mathrm{t}=\mathrm{x} / 2 \text { we get } \mathrm{y}=4 \mathrm{xx}^2 / 4 \\
& \mathrm{y}=\mathrm{x}^2 \\
& X=2 t \\
& Y=4 t^2 \\
& Y=x^2
\end{aligned}
$

Hence, the answer is the option (1).

Example 4: A projectile is given an initial velocity of $(\hat{i}+2 \hat{j}) \mathrm{m} / \mathrm{s}$, where $\hat{i}$ is along the ground and $\hat{j}$ is along the vertical. If $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$, the equation of its trajectory is :

1) $4 y=2 x-25 x^2$
2) $y=x-5 x^2$
3) $y=2 x-5 x^2$
4) $4 y=2 x-5 x^2$

Solution:

Given :
$
\vec{u}=\hat{i}+2 \hat{j}
$

From the equation of motion,
$
\begin{aligned}
& \vec{s}=\vec{u} t+\frac{1}{2} a t^2 \\
& \Rightarrow \overrightarrow{r_f}-\overrightarrow{r_i}=(\hat{i}+2 \hat{j}) t+\frac{1}{2}(-g \hat{j}) t^2 \\
& \Rightarrow \overrightarrow{r_f}-0=t \hat{i}+2 t \hat{j}-\frac{g}{2} t^2 \hat{j} \\
& \Rightarrow x \hat{i}+y \hat{j}=t \hat{i}+\left(2 t-\frac{g}{2} t^2\right) \hat{j}
\end{aligned}
$

After comparing both sides, we get,
$
x=t \text { and } y=\left(2 t-5 t^2\right)
$

Now, put the value of $\mathrm{t}$ in $y=\left(2 t-5 t^2\right)$

$\text{Note: the above question is solved by the basic method of kinematical equation motion but you can also solve this question by the equation of trajectory of projectile motion. Hint for alternate method :}$
$
y=x \tan \theta\left(1-\frac{x}{R}\right)
$
$\mathrm{OR}$,
$
y=x \tan \theta-\frac{1}{2} \frac{x^2 g}{u^2 \cos ^2 \theta}
$

Hence, the answer is the option (3).

Example 5: A body of mass $10 \mathrm{~kg}$ is projected at an angle of $45^{\circ}$ with the horizontal. The trajectory of the body is observed to pass through a point $(20,10)$. If $\mathrm{T}$ is the time of flight, then its momentum vector, at time $\mathrm{t}=\frac{\mathrm{T}}{\sqrt{2}}$ $\qquad$
$\left[\right.$ Take $\left.g=10 \mathrm{~m} / \mathrm{s}^2\right]$
$\left[\text { Take } g=10 \mathrm{~m} / \mathrm{s}^2\right]$

1) $100 \mathrm{i}+(100 \sqrt{2}-200) \mathrm{j}$
2) $100 \sqrt{2} \hat{\mathrm{i}}+(100-200 \sqrt{2}) \hat{\mathrm{j}}$
3) $100 \hat{\mathrm{i}}+(100-200 \sqrt{2}) \hat{\mathrm{j}}$
4) $100 \sqrt{2} \hat{\mathrm{i}}+(100 \sqrt{2}-200) \hat{\mathrm{j}}$

Solution:

$
\mathrm{m}=10 \mathrm{~kg}
$

We know that the equation of trajectory is
$
\begin{aligned}
& \mathrm{y}=\mathrm{x} \tan \theta\left(1-\frac{\mathrm{x}}{\mathrm{R}}\right) \\
& 10=20\left(\tan 45^{\circ}\right)\left(1-\frac{20}{\mathrm{R}}\right) \\
& \frac{1}{2}=1-\frac{20}{\mathrm{R}} \\
& \frac{1}{2}=\frac{20}{\mathrm{R}} \\
& \mathrm{R}=40 \rightarrow(1)
\end{aligned}
$

Also, we know that,

$\begin{aligned}
& \mathrm{R}=\frac{\mathrm{u}^2 \sin 2 \theta}{\mathrm{g}} \\
& 40=\frac{\mathrm{u}^2 \sin 90^{\circ}}{10} \\
& \mathrm{u}^2=400 \\
& \mathrm{u}=20 \mathrm{~m} / \mathrm{s} \rightarrow(2) \\
& \text { At } \mathrm{t}=\frac{\mathrm{T}}{\sqrt{2}} \\
& \overline{\mathrm{V}}=\overline{\mathrm{v}}_{\mathrm{x}}+\overline{\mathrm{V}}_{\mathrm{y}} \\
& =\mathrm{u} \cos \theta(\hat{\mathrm{i}})+(\mathrm{u} \sin \theta-\mathrm{g}) \hat{\jmath} \\
& =10 \sqrt{2} \hat{\mathrm{i}}+\left(10 \sqrt{2}-10 \times \frac{\mathrm{T}}{\sqrt{2}}\right) \hat{\jmath} \\
& \text { But } \mathrm{T}=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}=\frac{2 \times 20 \times}{10} \frac{1}{\sqrt{2}}
\end{aligned}$

$\begin{aligned}
& \mathrm{T}=2 \sqrt{2} \\
& \overline{\mathrm{v}}=10 \sqrt{2} \hat{\mathrm{i}}+(10 \sqrt{2}-20) \hat{\jmath} \\
& \therefore \text { Momentum vector }=\overline{\mathrm{P}}=\mathrm{m} \overline{\mathrm{v}} \\
& \overline{\mathrm{P}}=100 \sqrt{2} \hat{\mathrm{i}}+(100 \sqrt{2}-200) \hat{\mathrm{j}}
\end{aligned}$

Hence, the answer is option (1).

Summary

In summary, the equation of the path of a projectile is one of the cornerstone equations of physics, permitting great precision in the prediction and, hence, analysis of such described trajectories of the objects thrown into the atmosphere. This general movement, forming a parabola, due to the known movements under the law of the earth's gravity force in action, is core to understanding and optimizing the trajectory of the moving projectiles in different real-life scenarios.