Two seconds after projection, a projectile is moving at 30° above the horizontal, after one more second, it is moving horizontally. The initial speed of the projectile is (Take g=10ms−2 )
Let v1 be the initial speed at ' O ' and ' θ ' be the initial angle of projection.
After two seconds particle reaches at Q and θ1=30°. Hence, v2‌cos‌30°=v1‌cos‌θ v2.
√3
2
=v1‌cos‌θ v2=
2
√3
v1‌cos‌θ ........(i) By equation of motion, when particle reaches from O to Q v2‌sin‌30°=v1‌sin‌θ−2g
v2
2
=v1‌sin‌θ−2g
1
2
.
2
√3
v1‌cos‌θ=v1‌sin‌θ−2g v1‌cos‌θ=√3v1‌sin‌θ−2√3g .........(ii) After 3 seconds, i.e. at top position, particle moves in horizontal direction, hence vertical component is zero. ∴0=v1‌sin‌θ−g×t 0=v1‌sin‌θ−3g orv1=
3g
sin‌θ
........(iii) Putting the value of v1 from Eq (iii) to Eq (ii), we have, 3g