Question

A stone projected from ground with certain speed at a angle @ With horizontal attains maximum height H1,when it is projected with same speed at an angle @ with vertical attains height H2.The horizontal range of projectile is .?

Answer 1

A stone is projected from ground with certain speed , u at an angle $\theta$ with horizontal attains maximum height $H_1$.
so, $H_1=\frac{u^2\sin^2\theta}{2g}$...(1)

but when stone is projected with same speed , u at angle $\theta$ with vertical attains maximum height $H_2$.
so, $H_2=\frac{u^2sin^2(90-\theta)}{2g}$

$H_2=\frac{u^2cos^2\theta}{2g}$...(2)

now, range in case 1 :-
$R_1=\frac{u^2sin2\theta}{g}$
range in case 2 :-
$R_2=\frac{u^2sin2(90-2\theta)}{g}\\\\=\frac{u^2sin(180-2\theta)}{g}\\\\=\frac{u^2sin2\theta}{g}$

here, $R_1=R_2=R(\textbf{let})$

now from equations (1) and (2),
$H_1.H_2=\frac{u^4sin^2\theta.cos^2\theta}{4g^2}$

$H_1.H_2=\frac{u^4(2sin\theta.cos\theta)^2}{16g^2}$

taking square root both sides,

$\sqrt{H_1.H_2}=\frac{u^2sin2\theta}{4g}$

$\sqrt{H_1.H_2}=\frac{R}{4}$

or, $R=4\sqrt{H_1H_2}$

Answer 2

Answer:

4√h1h2

Explanation:

if a ball projected at two complementary angle with same velocity than horizontal range will be same