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?

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 should the particle be aimed to have range equal to maximum height? (A) H (B) 2 H (C) H / 2 (D) 3 H

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/63b39d38d76f0ee44835040f38e5ea6a-.png" /> A C=R / 2, P C=H We have to find h=M P. From the previous solution, we know that if H=R, then tan θ=4 Now tan θ=(M C/A C)=(M P+P C/A C) 4 =(h+H/R / 2) ⇒ 4=((h+H) 2/H) ⇒ h =H

Q. 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?

$H$

$2 H$

$H /2$

$3 H$

$A C = R /2, PC = H$
We have to find $h = MP$ .
From the previous solution, we know that if $H = R$ , then
$tan θ = 4$
Now $tan θ = \frac{MC}{A C} = \frac{MP + PC}{A C}$
$4 = \frac{h + H}{R /2} \Rightarrow 4 = \frac{( h + H ) 2}{H}$
$\Rightarrow h = H$

Q. 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?

H

2H

H/2

3H

AC=R/2,PC=H We have to find h=MP. From the previous solution, we know that if H=R, then tanθ=4 Now tanθ=ACMC​=ACMP+PC​ 4=R/2h+H​⇒4=H(h+H)2​ ⇒h=H

Q. 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?

$H$

$2 H$

$H /2$

$3 H$

$A C = R /2, PC = H$
We have to find $h = MP$ .
From the previous solution, we know that if $H = R$ , then
$tan θ = 4$
Now $tan θ = \frac{MC}{A C} = \frac{MP + PC}{A C}$
$4 = \frac{h + H}{R /2} \Rightarrow 4 = \frac{( h + H ) 2}{H}$
$\Rightarrow h = H$