Question

the horizontal rays of a projectile is 2√3 times of it's maximum hight find angle of projection ?

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Answer 1

Answer:

HEY MATEE✨✨

Explanation:

If u and α are the initial velocity of projection and angle of projection, respectively, then

$Maximum \: height \: attained =\frac{u {}^{2}sin {}^{2} \alpha }{2g}$

$Horizontal \: range =\frac{2u {}^{2}sin {}^{2} \alpha \cos \alpha }{g}$

According to the problem,

$\frac{2u {}^{2}sin {}^{2} \alpha \cos \alpha }{g} = 2 \: \sqrt[]{3} ( \frac{u {}^{2}sin {}^{2} \alpha }{2g} )$

=> tan α = tan-¹ (2/√3)

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