Question

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..
1. 8u^2/17g 2. 5 u^2/17g 3. 6 u^2 / 13g 4. u^2/g

Answer 1

let the angle made by the projectile from horizontal b 'x'
equate the max ht. and horizontal range...u will find a numerical vaue of x...put it back in the expreaaion of horizontal range...and u will get the ans..

Answer 2

I have tried that way but my answer is coming U^2/g which is wqrong one . The correct answer as per answer sheet is 8 u^2 /17g. Pl solve it ..

Answer 3

let the angle be A
maximum height be H initial velocity =u
v^=u^2-2gh
at highest position v=0
so ...0=(usinA)^2-2gH
or H=(usinA)^2/2g

range R=ucosA*t
R=ucosA*2usinA/g [t=2usinA/g]

now H=R
(usinA)^2/2g=2u^2sinA*cosA/g
or sinA=4cosA
or tanA=4
A=tan^-14

now the range R=2u^2/g*sin(tan^-14)*cos(tan^-14) =8u^2/(17g)

Answer 4

Thnx Souvik nd RaGhavv