A stone projected with a speed u at an angle - with the horizontal reaches maximum height H 1 - When it is projected with speed u at an angle 90- with the horizontal- it reaches maximum height H 2 - The relation between the horizontal range R of the projectile- H 1 and H 2 is A- R-4 -H1 H2B- R -4 H 1- H 2C- R-4H1-H2D - R-H12-H22

The correct option is A H_1 = u^2sin^2θ/2g and H_2 = u^2sin^2 (90 θ)/2g = u^2cos^2θ/2g H_1 H_2 = u^2sin^2θ/2g × u^2cos^2θ/2g = (u^2 sin 2 θ)^2/16g^2= R^2/16 ...

Byju's Answer

Standard XII

Physics

Projectile Time, Height and Range

A stone proje...

Question

A stone projected with a speed u at an angle θ with the horizontal reaches maximum height $H_{1}$ . When it is projected with speed u at an angle $90 - θ$ with the horizontal, it reaches maximum height $H_{2}$ . The relation between the horizontal range R of the projectile, $H_{1}$ and $H_{2}$ is

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R = 4(H₁ - H₂)

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Solution

The correct option is A
$H_{1}$ = $\frac{u^{2} {s i n}^{2} θ}{2 g}$ and
$H_{2}$ = $\frac{u^{2} {s i n}^{2} (90 - θ)}{2 g}$ = $\frac{u^{2} {c o s}^{2} θ}{2 g}$
$H_{1} H_{2}$ = $\frac{u^{2} {s i n}^{2} θ}{2 g}$ $\times$ $\frac{u^{2} {c o s}^{2} θ}{2 g}$ = $\frac{{(u^{2} s i n 2 θ)}^{2}}{16 g^{2}}$ = $\frac{R^{2}}{16}$
$∴$ R = 4 $\sqrt{H_{1} H_{2}}$

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