Question

One second after projection, a stone moves at an angle ${45}^{\circ }$ with horizontal. Two seconds after projection, it moves horizontally. It's angle of projection is-
$[g=10\frac{m}{s^2}]$

• A. $ta{n}^{-1}\left(\sqrt{3}\right)$
• B. $ta{n}^{-1}\left(4\right)$
• C. $ta{n}^{-1}\left(3\right)$
• D. $ta{n}^{-1}\left(2\right)$

Answer 1

The correct option is D $ta{n}^{-1}\left(2\right)$

Let the angle of projection be $\theta$.
Then the time taken to reach the maximum height where the velocity is horizontal, $T=\frac{usin\theta }{g}=2s$
Hence $usin\theta =2g$-----------(1)
Now after one second the projectile makes an angle of ${45}^{\circ }$.
Hence the vertical component of velocity will be equal to horizontal component of velocity there. We know that the horizontal component of velocity remains constant in projectile which is $ucos\theta$
Hence the vertical component of velocity will also be $ucos\theta$.
Applying equation of motion after 1 sec.
$v_y=u_y+a\times t$
$ucos\theta =usin\theta -g\times 1$
From equation 1 we know $usin\theta =2g$
Putting this value in the above equation
$ucos\theta =2g-g=g$--------------(2)
From equation 1 and 2
$tan\theta =2$
or $\theta =ta{n}^{-1}2$