A projectile is thrown with a velocity of 60m/s in a direction making an angle of 30° with horizontal.find it's velocity after2s
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horizontal component of velocity=60×cos30°=60×√3/2=30√3 and it will remain constant as no acceleration in horizontal direction.
vertical component=60sin30°=60×1/2=30
vertical acceleration =-g=-10
vertical velocity after 2 sec =30-10×2
[using v=u+at,all symbols has normal meaning]
vertical velocity after 2 sec=30-20=10
so after 2 sec
vertical velocity=10 upward direction
horizontal velocity=30√3
net velocity =√[(10)^2+(30√3)^2]
=√[100+2700]
=√[2800]
=20√7
vertical component=60sin30°=60×1/2=30
vertical acceleration =-g=-10
vertical velocity after 2 sec =30-10×2
[using v=u+at,all symbols has normal meaning]
vertical velocity after 2 sec=30-20=10
so after 2 sec
vertical velocity=10 upward direction
horizontal velocity=30√3
net velocity =√[(10)^2+(30√3)^2]
=√[100+2700]
=√[2800]
=20√7
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