JEE MAIN - Physics (2022 - 27th June Morning Shift - No. 1)
A projectile is launched at an angle '$$\alpha$$' with the horizontal with a velocity 20 ms$$-$$1. After 10 s, its inclination with horizontal is '$$\beta$$'. The value of tan$$\beta$$ will be : (g = 10 ms$$-$$2).
tan$$\alpha$$ + 5sec$$\alpha$$
tan$$\alpha$$ $$-$$ 5sec$$\alpha$$
2tan$$\alpha$$ $$-$$ 5sec$$\alpha$$
2tan$$\alpha$$ $$+$$ 5sec$$\alpha$$
Explanation
At $t=0$, the motion of projectile is given as
_27th_June_Morning_Shift_en_1_1.png)
$\tan \alpha=\frac{u_{y}}{u_{x}}$
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
_27th_June_Morning_Shift_en_1_2.png)
$\tan \beta=\frac{v_{y}}{v_{x}}$ .........(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
$$ \begin{aligned} &v_{y}=u_{y}-g t \\\\ &v_{y}=u_{y}-100 \end{aligned} $$
Using Eq, (iii)}
$$ \frac{\tan \beta}{\tan \alpha}=\frac{u_{y}-100}{u_{y}} $$
$$ \begin{aligned} & \frac{\tan \beta}{\tan \alpha}=1-\frac{100}{u_{y}}=1-\frac{100}{20 \sin \alpha} \quad\left(\because u_{y}=20 \sin \alpha\right) \end{aligned} $$
$$ =1-\frac{5}{\sin \alpha} $$
$\Rightarrow \tan \beta=\tan \alpha\left(1-\frac{5}{\sin \alpha}\right)=\tan \alpha-5 \sec \alpha$
$\tan \alpha=\frac{u_{y}}{u_{x}}$
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
$$ \begin{aligned} &v_{y}=u_{y}-g t \\\\ &v_{y}=u_{y}-100 \end{aligned} $$
Using Eq, (iii)}
$$ \frac{\tan \beta}{\tan \alpha}=\frac{u_{y}-100}{u_{y}} $$
$$ \begin{aligned} & \frac{\tan \beta}{\tan \alpha}=1-\frac{100}{u_{y}}=1-\frac{100}{20 \sin \alpha} \quad\left(\because u_{y}=20 \sin \alpha\right) \end{aligned} $$
$$ =1-\frac{5}{\sin \alpha} $$
$\Rightarrow \tan \beta=\tan \alpha\left(1-\frac{5}{\sin \alpha}\right)=\tan \alpha-5 \sec \alpha$
Comments (0)
