Physics - Kinematics Question with Solution | TestHub

Let in 2s body reaches upto point A and after one more second upto point B.

Total time of ascent for a body is given as 3s



i.e.,$\frac{u\sin \theta }{g}=3$

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

Horizontal component of velocity remains always constant

$u\cos \theta =v\cos 30°$.......(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$$\left[As u\sin \theta =30\right]$

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

Substituting this value$v$in equation (ii), we get

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

From equations (i) and (ii)

$u=20\sqrt{3}$${\text{m s}}^{-1}$and$\theta =60°$