Question

One second after projection, a stone moves at an angle $45$ with horizontal. Two seconds after projection, it moves horizontally, its angle of projection is- $[$$g=10m/s^2$$]$

• 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 $ta{n}^{-1}\left(2\right)$

when the stone moves horizontally it means it is at the top most point of motion

therefore time of flight is 2s hence half of time of flight given as 1 s therefore

$v=u+at$

when stone is at top

$0=v_y-g\times 2$

$v_y=2g$

after 1 s of projection

$v=v_y-g\times 1=g$

as the angle is 45 at 1s after projection therefore horizontal and vertical component are both equal i.e. $v_y=v_x=10m/s$

As we know horizontal component alwats remains same therefore angle of projection is

$tan\theta =\frac{v_y}{v_x}=\frac{20}{10}=2$

$\theta =ta{n}^{-1}\left(2\right)$