Question

The position of a projectile launched from the origin at $t=0$ is given by $\vec{r}=\left( 40\hat{i}+50\hat{j} \right)m$ at $t=2s$. If the projectile was launched at an angle $\theta$ from the horizontal, then $\theta$ is (take $g=10m{{s}^{-2}}$).
$\begin{aligned} & \text{A}\text{. }{{\tan }^{-1}}\dfrac{2}{3} \\\ & \text{B}\text{. }{{\tan }^{-1}}\dfrac{3}{2} \\\ & \text{C}\text{. }{{\tan }^{-1}}\dfrac{7}{4} \\\ & \text{D}\text{. }{{\tan }^{-1}}\dfrac{4}{5} \\\ \end{aligned}$

Answer 1

To solve this question we need to use the concept of projectile motion of a particle. We are given the position of the particle at two different times. Express the vertical and the horizontal velocity for the projectile motion and find the vertical distance and the horizontal distance separately in terms of the components of the velocity,then take the ratio to find the required angle.

Complete answer:
The projectile is launched from the origin at $t=0$.
Considering the x-y coordinate system, at $t=0$, $x=0,y=0$
The position of a projectile is given by $\vec{r}=\left( 40\hat{i}+50\hat{j} \right)m$ at $t=2s$.
At $t=2s$, $x=40, y=50$
The projectile is launched at angle $\theta$ with the horizontal.
Let, the initial velocity of the projectile is u.
The horizontal component of the velocity of the projectile will be $u\cos \theta$
The vertical component of velocity of the projectile will be $u\sin \theta$
So, after 2 sec, the horizontal distance covered by the object will be,
$\begin{aligned} & u\cos \theta t=40 \\\ & u\cos \theta \times 2=40 \\\ & u\cos \theta =20 \\\ \end{aligned}$
Again, after 2 sec, the vertical distance covered will be,
$\begin{aligned} & u\sin \theta t-\dfrac{1}{2}g{{t}^{2}}=50 \\\ & u\sin \theta \times 2-\dfrac{1}{2}\times 10\times {{2}^{2}}=50 \\\ & 2u\sin \theta -20=50 \\\ & 2u\sin \theta =70 \\\ & u\sin \theta =35 \\\ \end{aligned}$
Dividing the second equation by the first equation, we get that,
$\begin{aligned} & \dfrac{u\sin \theta }{u\cos \theta }=\dfrac{35}{20} \\\ & \tan \theta =\dfrac{7}{4} \\\ & \theta ={{\tan }^{-1}}\dfrac{7}{4} \\\ \end{aligned}$
So, the projectile was launched at an angle of ${{\tan }^{-1}}\dfrac{7}{4}$ from the horizontal direction.

The correct option is (C).

Note:
The projectile motion of an object is due to the force of gravity on an object thrown into space. When we take the horizontal component of the velocity and we find the horizontal distance, there is no action due to the force of gravity. But, when we take the vertical component of the velocity and then find the vertical distance covered, we need to take the action of the gravity on the projectile.