QJEEM22Ph0aed: JEE Main 2022 Physics - Class 11 JEEM22Ph0aed Solution: A projectile is launched at an angle ' α ' with the horizontal…

At $t=0$ , the motion of projectile is given as



$\tan \alpha=\frac{u_{y}}{u_{x}}$

where, $u_{y}$ is the vertical component of initial velocity and $u_{x}$ is the horizontal component of initial velocity.

At $t=10 \mathrm{~s}$ , the motion of projectile is given as



$\tan \beta=\frac{v_{y}}{v_{x}}$ .........(ii)

where, $v_{y}$ is the vertical component of final velocity after $t=10 \mathrm{~s}$ and $v_{x}$ is the horizontal component of final velocity.

From Eqs. (i) and (ii), we get $\frac{\tan \beta}{\tan \alpha}=\frac{v_y}{u_y} \quad\left(\right.$ as $v_x=u_x$ ) ....(iii)

Using the following equation in vertical direction, we get

$$\begin{array}{rl}& v_y=u_y-gt\\ \\ & v_y=u_y-100\end{array}$$

Using Eq, (iii)}

$$\frac{\tan \beta }{\tan \alpha }=\frac{u_y-100}{u_y}$$

$$\begin{array}{rl}& \frac{\tan \beta }{\tan \alpha }=1-\frac{100}{u_y}=1-\frac{100}{20\sin \alpha }(\because u_y=20\sin \alpha )\end{array}$$

$$=1-\frac{5}{\sin \alpha }$$

$\Rightarrow \tan \beta=\tan \alpha\left(1-\frac{5}{\sin \alpha}\right)=\tan \alpha-5 \sec \alpha$