Manipal Entrance Test 2014 Physics Paper

Let in 2 s 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.e., t=

u‌sin‌θ

g

=3
∴ u‌sin‌θ=10×3
u‌sin‌θ=30 .....(i)
Horizontal component of velocity remains always constant
u‌cos‌θ=v‌cos‌30°....(ii)
For vertical upward motion between point O and A
v‌sin‌30°=u‌sin‌θ−g×2‌‌‌ [Using‌v=u−gt]
v‌sin‌30°=30−20 [ As u‌sin‌θ=30]
∴ v=20m/s
Substituting this value v, we get in Eq. (ii)
u‌cos‌θ=20‌cos‌30°=10√3 .....(iii)
From Eqs. (i) and (iii)
u=20√3 and θ=60°