Question

Two bodies are thrown with the same initial velocity at angle alpha and 90- alpha with the horizontal what will be the ratio of i) maximum height attained by them ii) of horizontal range​

Answer 1

Explanation:

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Answer 2

Answer:

(i) $sin^{2} : cos^{2}$

(ii) 1:1

Explanation:

Let the 2 bodies be thrown with initial velocity 'u'.

We know

Range = $\frac{u^{2}sin2\alpha }{g}$

Height = $\frac{u^{2}sin^{2}\alpha }{2g}$

For first body:

Range = $R_{1}$ = $\frac{u^{2}sin2\alpha }{g}$ ____(1)

Height = $H_{1} = \frac{u^{2}sin^{2}\alpha }{2g}$ _____(2)

For second body:

Range = $R_{2} = \frac{u^{2}sin2\ (90-\alpha) }{g}$ = $\frac{u^{2}sin\ (180-2\alpha) }{g} = \frac{u^{2}sin2\alpha }{g}$ _____(3)

Height = $H_{2} = \frac{u^{2}sin^{2}\ (90-\alpha ) }{2g}$ = $\frac{u^{2}cos^{2}\ \alpha }{2g}$ _____(4)

(i) Ratio of maximum height

$\frac{H_{1} }{H_2} = \frac{u^{2}sin^{2}\alpha }{2g} . \frac{2g}{u^{2}cos^{2}\alpha }$

= $\frac{H_1}{H_2} = \frac{sin^{2}\alpha }{cos^{2}\alpha } \\H_{1} : H_2 = sin^{2} \alpha : cos^{2} \alpha$

(ii) Ratio of horizontal range

$\frac{R_1}{R_2} = \frac{u^{2}sin2\alpha }{g} . \frac{g}{u^{2} sin2\alpha }$

$\frac{R_1}{R_2} = \frac{1}{1}$

$R_1:R_2 = 1:1$