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A particle is projected with a velocity v=(3i-j+2k) and a constant acceleration acting on the particle is a=(-6i+2j-4k) then path of projectile is
Asked by rawatvaishali2020 | 26 Nov, 2020, 10:21: PM
Path of projectile is determined using the equation of motion S = u t + (1/2) a t2 .................. (1)

where u is initial speed and a is acceleration

we are given :-    = ( 3  + 2  )   ........................(2)
and    = ( -6 +2  -4  )  ......................(3)
By substituting u and a in eqn.(1) , we get

= ( 3  + 2  ) t +(1/2) ( -6 +2  -4  ) t2
By simplifying above expression , we get , S = ( t - t2 ) ( 3  + 2  ) ..............(4)
if we compare eqn.(4) and (2) , we get ,  = ( t - t2 )
Answered by Thiyagarajan K | 26 Nov, 2020, 10:54: PM
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