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        A body is projected with velocity of 60 m/s at an angle of 300 with the horizantal . Find horizantal distance covered by the projectile in first two seconds
one year ago

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

one year 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* √3/2Since acceleration along x direction is 0 there will be no change in velocitySo Sx ( displacement along x direction ) = Ux *t  = 30√3 * 2 = 60√3 m.

one year 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............................

one year ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions