Question

A body is projected with a certain speed at angles of projection of theta and 90-theta ,the maximum height attained in two cases are 20m,10m respectively.The maximum possible range is?

Answer 1

maximum height of projectile is given by equation

$H_1 = \frac{v^2 sin^2\theta}{2g}$

$H_2 = \frac{v^2sin^2(90-\theta)}{2g}$

given here that

$H_1 = 20 m$

$H_2 = 10 m$

now if we divide two equations then we have

$\frac{20}{10} = \frac{sin^2\theta}{sin^2(90-\theta)}$

$tan\theta = \sqrt2$

now we have

$sin^2\theta = \frac{2}{3}$

now we have

$H = \frac{v^2sin^2\theta}{2g}$

$20 = \frac{v^2*\frac{2}{3}}{2g}$

maximum range is possible at 45 degree which is given by

$R = \frac{v^2}{g}$

now from the above equation we can say

$R = \frac{v^2}{g} = 60 m$

Answer 2

Answer:

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Explanation:

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