Two second after projection, a projectile is travelling in a direction inclined at 300 with the horizontal.After 1 more second, it is travelling horizontally. Then, (Take, g = 10 ms-2)

T/2=2+1=3 s or T=6 s⇒ 2uyg=6∴ uy=30 ms-1Further, tan30°=vyvx=uy-gtux=30-20uxor ux=103 ms-1or u=ux2+uy2=203 ms-1

Two second after projection, a projectile is travelling in a direction inclined at 300 with the horizontal.After 1 more second, it is travelling horizontally. Then, (Take, g = 10 ms-2)

T/2=2+1=3 s or T=6 s⇒ 2uyg=6∴ uy=30 ms-1Further, tan30°=vyvx=uy-gtux=30-20uxor ux=103 ms-1or u=ux2+uy2=203 ms-1

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Two second after projection, a projectile is travelling in a direction inclined at 30⁰ with the horizontal.
After 1 more second, it is travelling horizontally. Then, (Take, g = 10 ms^-2)

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the angle of projection is 30⁰ with horizontal

Both (1) and (2) are incorrect

Both (1) and (2) are correct

the velocity of projection is 20 $\sqrt{3} m s^{- 1}$

answer is A.

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$T / 2 = 2 + 1 = 3 s or T = 6 s$

$\Rightarrow \frac{2 u_{y}}{g} = 6$

$∴ u_{y} = 30 {ms}^{- 1}$

Further, $\tan 30 ° = \frac{v_{y}}{v_{x}} = \frac{u_{y} - g t}{u_{x}} = \frac{30 - 20}{u_{x}}$

or $u_{x} = 10 \sqrt{3} {ms}^{- 1}$

or $u = \sqrt{u_{x}^{2} + u_{y}^{2}} = 20 \sqrt{3} {ms}^{- 1}$

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Two second after projection, a projectile is travelling in a direction inclined at 300 with the horizontal.After 1 more second, it is travelling horizontally. Then, (Take, g = 10 ms-2)

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Two second after projection, a projectile is travelling in a direction inclined at 30⁰ with the horizontal.
After 1 more second, it is travelling horizontally. Then, (Take, g = 10 ms^-2)