A particle is projected from a point on ground. It just clears a vertical wall of height h at a distance h from the point of projection and finally hits the ground at a distance 3h from the base of the wall. Then maximum height attained by the projectile is
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answer is 1.
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Detailed Solution
y=xtanα1-xR∴h=htanα1-h4h⇒tanα=43At x=R2=2h, Height (y) attained by the projectile is maximum.∴H=ymax=2h.431-2h4h=4h3