Question 73: Motion in 2D & Vectors

Question

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).

Accepted Answer

At $t=0$, the motion of projectile is given as
$	an \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
$	an \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 $\frac{	an \beta}{	an \alpha}=\frac{v_y}{u_y} \quad\left(ight.$ 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)}
$$
\frac{	an \beta}{	an \alpha}=\frac{u_{y}-100}{u_{y}}
$$
$$
\begin{aligned}
& \frac{	an \beta}{	an \alpha}=1-\frac{100}{u_{y}}=1-\frac{100}{20 \sin \alpha} \quad\left(\because u_{y}=20 \sin \alphaight)
\end{aligned}
$$
$$
=1-\frac{5}{\sin \alpha}
$$
$\Rightarrow 	an \beta=	an \alpha\left(1-\frac{5}{\sin \alpha}ight)=	an \alpha-5 \sec \alpha$

Answer

tan$$\alpha$$ + 5sec$$\alpha$$

Answer

tan$$\alpha$$ $$-$$ 5sec$$\alpha$$

Answer

2tan$$\alpha$$ $$-$$ 5sec$$\alpha$$

Answer

2tan$$\alpha$$ $$+$$ 5sec$$\alpha$$

Motion in 2D & Vectors

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).

Choose the correct answer: