The position of a projectile launched from the origin at \( t=0 \) is given by \( \vec{r}=(40 \hat{ i }+50 \hat{ j }) m \) at \( t=2 s \). If the projectile was launched at an angle \( \theta \) from the horizontal, then \( \theta \) is (take \( g=10 ms^{-2} \) ) [JEE Main 2019] (1) \( \tan ^{-1} \frac{2}{3} \) (2) \( \tan ^{-1} \frac{3}{2} \) (3) \( \tan ^{-1} \frac{7}{4} \) (4) \( \tan ^{-1} \frac{4}{5} \)

Question

The position of a projectile launched from the origin at \( t=0 \) is given by \( \vec{r}=(40 \hat{ i }+50 \hat{ j }) m \) at \( t=2 s \). If the projectile was launched at an angle \( \theta \) from the horizontal, then \( \theta \) is (take \( g=10 ms^{-2} \) )

(1) \( \tan ^{-1} \frac{2}{3} \)

 (2) \( \tan ^{-1} \frac{3}{2} \) 

(3) \( \tan ^{-1} \frac{7}{4} \) 

(4) \( \tan ^{-1} \frac{4}{5} \)

Answer 1

Correct option is (3) \(\tan^{-1} \frac 74\)

As we know that horizontal component of projected velocity remains constant.

So, from question,

Horizontal velocity, u_x = \(\frac{40 - 0}{2-0} = 20 m/s\)

And, initial vertical velocity (u_y),

\(S_y= u_yt + \frac 12 at^2\)

\(50 = u_y(2) - \frac 12 (10)(2)^2\)

\(\Rightarrow u_y = \frac{70}2 = 35 m/s\)

\(\therefore \tan \theta = \frac{u_y}{u_x} = \frac{35}{20} = \frac 74\)

\(\Rightarrow\) Angle, \(\theta = \tan^{-1} \frac 74\)