Question

Two seconds after projection, a projectile is travelling in a direction inclined at ${30}^o$ to the horizontal. After one more second, it is travelling horizontally. Find the magnitude and direction of its velocity.

Answer 1

let in 2 sec body reaches upto point $A$ and after one more sec upto point $B$

Total time of ascent for a body is given $3$s ,i.,

$t=\frac{u\sin \theta }{g}=3$

$\therefore u\sin \theta =10\times 3=30$ .........(i)

Horizontal component of velocity remains always constant,

$u\cos \theta =v\cos {30}^o$..........................(ii)

For vertical upward motion between point $O$ and $A$.

$v\sin {30}^o=u\sin \theta -g\times 2$ [Using $v=u-gt$]

$v\sin {30}^o=30-20$ [As $u\sin \theta =30$]

$\therefore v=20m{s}^{-1}$

Substituting this value in Eq.(ii),

$u\cos \theta =20\cos {30}^o=10\sqrt{3}$ ...........................(iii)

From eq (i) and (iii) , $u=20\sqrt{3}$ and $\theta ={60}^o$.