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CBSE Class 11-science Answered

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, 22:21: PM
answered-by-expert Expert Answer
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 :-   begin mathsize 14px style u with rightwards arrow on top end style = ( 3 begin mathsize 14px style i with hat on top end style- begin mathsize 14px style j with hat on top end style + 2 begin mathsize 14px style k with hat on top end style )   ........................(2)
   and   begin mathsize 14px style a with rightwards arrow on top end style = ( -6 begin mathsize 14px style i with hat on top end style+2 begin mathsize 14px style j with hat on top end style -4 begin mathsize 14px style k with hat on top end style )  ......................(3)
By substituting u and a in eqn.(1) , we get 
 
begin mathsize 14px style S with rightwards arrow on top end style = ( 3 begin mathsize 14px style i with hat on top end style- begin mathsize 14px style j with hat on top end style + 2 begin mathsize 14px style k with hat on top end style ) t +(1/2) ( -6 begin mathsize 14px style i with hat on top end style+2 begin mathsize 14px style j with hat on top end style -4 begin mathsize 14px style k with hat on top end style ) t2 
By simplifying above expression , we get , S = ( t - t2 ) ( 3 begin mathsize 14px style i with hat on top end style- begin mathsize 14px style j with hat on top end style + 2 begin mathsize 14px style k with hat on top end style ) ..............(4)
if we compare eqn.(4) and (2) , we get , begin mathsize 14px style S with rightwards arrow on top end style = ( t - t2 ) begin mathsize 14px style u with rightwards arrow on top end style
Answered by Thiyagarajan K | 26 Nov, 2020, 22:54: PM

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