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Grade: 11
        
A body is projected with velocity of 60 m/s at an angle of 30with the horizantal . Find horizantal distance covered by the projectile in first two seconds
11 months ago

Answers : (3)

Manda Snehashish Reddy
113 Points
							
Angle is 30deg(let it be a), u=60m/s. We know that, horizontal distance depends on the horizontal component of velocity only and also the fact that is not affected by gravity. So, s=ucosa x t
      s=60 x cos30 x 2
      s=60 x 3^1/2
      s=103.923048454 metres
11 months ago
Piyush Kumar Maurya
65 Points
							
Since we have to calculate the distance travelled on x direction only. We will take all terms in x direction only. 
Vx = Vcos30 = 60* √3/2
Since acceleration along x direction is 0 there will be no change in velocity
So Sx ( displacement along x direction ) = Ux *t  = 30√3 * 2 = 60√3 m.
11 months ago
Khimraj
3008 Points
							
horizontal component of velocity = 60cos30 = 30\sqrt{3}
so distance covered in first two seconds = 2*30\sqrt{3} = 60\sqrt{3}
hOPE IT CLEARS............................
11 months ago
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