What are the special Points of Oblique Projectile Motion
What are the special Points of Oblique Projectile Motion?

Special Points of Oblique Projectile Motion :



(1) The three basic equations of motion, i.e.

$v=u+a t$        $s=u t+\frac{1}{2} a t^{2}$      $v^{2}=u^{2}+2 a s$

For projectile motion give :

$\mathrm{T}=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}$  $R=\frac{u^{2} \sin 2 \theta}{g}$    $H=\frac{u^{2} \sin ^{2} \theta}{2 g}$

(2) In the case of projectile motion,

The horizontal component of velocity $(u \cos \theta)$, acceleration $(g)$ and mechanical energy remains constant.

Speed, velocity, the vertical component of velocity (u $\sin \theta$ ), momentum, kinetic energy, and potential energy all change. Velocity and K.E. are maximum at the point of projection, while minimum (but not zero) at the highest point.

points of oblique projectile motion1 (3) If angle of projection is changed from

$\theta \stackrel{\text { to }}{\longrightarrow} \theta^{\prime}=(90-\theta)$

then range $\quad R^{\prime}=\frac{u^{2} \sin 2 \theta^{\prime}}{g}=\frac{u^{2} \sin 2(90-\theta)}{g}=\frac{u^{2} \sin 2 \theta}{g}=R$

points of oblique projectile motion2

So a projectile has same range for angles of projection $\theta$ and $(90-\theta)$

But has different time of flight $(\mathrm{T})$, maximum height $(\mathrm{H}) \&$ trajectories

Range is also same for $\theta_{1}=45^{\circ}-\alpha \quad$ and $\quad \theta_{2}=45^{\circ}+\alpha .\left[\right.$ equal $\left.\frac{u^{2} \cos 2 \alpha}{g}\right]$

(4) For maximum Range

$R=R_{\max } \Rightarrow 2 \theta=90^{\circ}$

for $\quad \theta=45^{\circ}$

$R_{\max }=\frac{u^{2}}{g}$  [For $\sin 2 \theta=1=\sin 90^{\circ}$ or $\theta=45^{\circ}$ ]

When range is maximum $\Rightarrow$ Then maximum height reached

$\mathrm{H}=\frac{\mathrm{u}^{2} \sin ^{2} 45}{2 \mathrm{~g}}\left(\right.$ When $\left.\mathrm{R}_{\max }\right) \quad$ or $\quad \mathrm{H}=\frac{\mathrm{u}^{2}}{4 \mathrm{~g}}$

hence maximum height reached (for $R_{\max }$ ) $\quad H=\frac{R_{\max }}{4}$

points of oblique projectile motion3

(5) For height $\mathrm{H}$ to be maximum

$\mathrm{H}=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}=\max \quad$ i.e. $\sin ^{2} \theta=1(\max )$ or for $\theta=90^{\circ}$

So that $\mathrm{H}_{\max }=\frac{\mathrm{u}^{2}}{2 \mathrm{~g}} \quad$ When projected vertically (i.e. at $\theta=90^{\circ}$ )

in this case Range $R=\frac{u^{2} \sin \left(2 \times 90^{\circ}\right)}{g}=\frac{u^{2} \sin 180^{\circ}}{g}=0$

$\mathrm{H}_{\max }=\frac{\mathrm{u}^{2}}{2 \mathrm{~g}}$ (For vertical projection) and $\mathrm{R}_{\max }=\frac{\mathrm{u}^{2}}{\mathrm{~g}}$ (For oblique projection with same velocity)

so $\quad \mathrm{H}_{\max }=\frac{\mathrm{R}_{\max }}{2}$

If a person can throw a projectile to a maximum distance $\left(w i t h, \theta=45^{\circ}\right) R_{\max }=\frac{u^{2}}{g}$.

The maximum height to which he can throw the projectile (with $\theta=90^{\circ}$ ) $\mathrm{H}_{\max }=\frac{\mathrm{R}_{\max }}{2}$

(6) At highest point

Potential energy will be max and equal to $(\mathrm{PE})_{\mathrm{H}}=\mathrm{mgH}=\mathrm{mg} \cdot \frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}$ or $(\mathrm{PE})_{\mathrm{H}}=\frac{1}{2} \mathrm{mu}^{2} \sin ^{2} \theta$.

While K.E. will be minimum (but not zero) and at the highest point as the vertical component of velocity is zero.

$(\mathrm{KE})_{\mathrm{H}}=\frac{1}{2} \mathrm{mv}_{\mathrm{H}}^{2}=\frac{1}{2} \mathrm{~m}(\mathrm{u} \cos \theta)^{2} \quad=\frac{1}{2} \mathrm{mu}^{2} \cos ^{2} \theta$

so $(\mathrm{PE})_{\mathrm{H}}+(\mathrm{KE})_{\mathrm{H}}=\frac{1}{2} \mathrm{mu}^{2} \sin ^{2} \theta+\frac{1}{2} \mathrm{mu}^{2} \cos ^{2} \theta=\frac{1}{2} \mathrm{mu}^{2}=$ Total M.E.

points of oblique projectile motion4 So in projectile motion mechanical energy is conserved.

$\left(\frac{P E}{K E}\right)_{H}=\frac{\frac{1}{2} m u^{2} \sin ^{2} \theta}{\frac{1}{2} m u^{2} \cos ^{2} \theta}=\tan ^{2} \theta$

(7) In case of projectile motion if range $R$ is $n$ times the maximum height $H$, i.e. $R=n H$

then $\frac{u^{2} \sin 2 \theta}{g}=n \cdot \frac{u^{2} \sin ^{2} \theta}{2 g}$

or $\quad 2 \cos \theta=\frac{n \cdot \sin \theta}{2}$

or $\quad \tan \theta=\frac{4}{n} \quad \Rightarrow \quad \theta=\tan ^{-1}\left(\frac{4}{n}\right)$

(8) Weight of a body in projectile motion is zero as it is a freely falling body.

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