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THe horizontal range of the projecile is R and the maximum height is H. A strong wind now blows in the direction of motion of the projectile giving it a constant horizontal acceleration =g/2 under the same conditions of projectile then find the range.

Purva , 6 Years ago
Grade 11
anser 1 Answers
Arun

Last Activity: 6 Years ago

Dear student
 
we know that in 1st  case  H=(u2sin2a)/2g   R=(u2sin2a)/g
now wright the equation of motion in second case
s=((ucosa)t-(1/2)(g/2)t2 )i-((usina)t-(1/2)gt2 )j
when partical lands on ground coffeicant of j =0 so t=(2usina)/g subatute above equation in cofficant of i to  get new range
so we get (u2sin2a)/g +(u2sin2a)/g
=R+2H
In case of any difficulty, please feel free to contact.
Regards
Arun (askIITians forum expert)

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