1from which bk r u asking this can u pls tell me
A body is projected with a speed u in such a direction that the maximum height obtained is equal to its horizontal range. The horizontal range is
a)8u2/17g
b)5u2/17g
c)6u2/13g
d)u2/2g
-
UP 0 DOWN 0 0 3
3 Answers
106max height obtained = u2sin2θ2g
horizontal range = 2usinθ*ucosθg
equating both, we get
sinθ/2 = 2cosθ
=> tanθ = 4
=> sin2θ = 16/17
=> max height = max range = u21617*2g= 8u2/17g
Raunaq Saha why is sin2θ = 16/17 ?Upvote·0· Reply ·2013-06-28 19:35:42
Himanshu Giria 1 + tan2 theta = sec2 theta. and. cos theta = 1 / sec theta. And. sin2 theta = 1 - cos 2 theta