Question

A particle is projected from ground at some angle with the horizontal. Let P be the point at maximum height H. At what height above the point P the particle should be aimed to have range equal to maximum height

Answer 1

max height = range

u2sin\Theta/2g  = 2u2sin\Thetacos\Theta/g

½ = 2cos\Theta

then

cos\Theta = ¼

then

tan\Theta = \sqrt{15}

let height above point is x then

tan\Theta  = (H+x)/(H/2) = \sqrt{15}

then x = (H/2)(\sqrt{15} – 2)

Hope it clears.

Answer 2

Answer:

Let P be the point at maximum height H. At what height above the point P the ... tan\Theta = (H+x)/(H/2) = \sqrt{ 15}.

max height = range u2sin\Theta/2g = 2u2sin\Thetacos\Theta/g ½ = 2cos\Theta then cos\Theta = ¼ then tan\Theta = \sqrt{15} let height above point .