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 $P$ should the particle be aimed to have range equal to maximum height?

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Solution

The correct option is D $H$
$A C = R / 2, P C = H$
we have to find $h = M P$
From the figure in previous question , we know that if $H = R$ then $tan θ = 4$
now $tan θ = \frac{M C}{A C} = \frac{M P + P C}{A C}$
$4 = \frac{h + H}{R / 2} \Rightarrow 4 = \frac{(h + H) 2}{H}$
$\Rightarrow h = H$

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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 should the particle be aimed to have range equal to maximum height?

The correct option is D HAC=R/2,PC=Hwe have to find h=MPFrom the figure in previous question , we know that if H=R then tanθ=4now tanθ=MCAC=MP+PCAC4=h+HR/2⇒4=(h+H)2H⇒h=H

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 $P$ should the particle be aimed to have range equal to maximum height?

The correct option is D $H$
$A C = R / 2, P C = H$
we have to find $h = M P$
From the figure in previous question , we know that if $H = R$ then $tan θ = 4$
now $tan θ = \frac{M C}{A C} = \frac{M P + P C}{A C}$
$4 = \frac{h + H}{R / 2} \Rightarrow 4 = \frac{(h + H) 2}{H}$
$\Rightarrow h = H$