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)