Question

find the angle of projection for which the horizontal range and the maximum height are equal

Answer 1

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GIVEN
Horizontal range= maximum height
or

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

$2 \sin( \alpha ) \cos( \alpha ) = \frac{ {sin}^{2} \alpha }{2}$

$\frac{sin \alpha }{cos \alpha } = 4$

$tan \: \alpha = 4$

alpha=75°58'

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Answer 2

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we have ...

♠ Maximum height = u^2 sin^2 tita/2g

♠ Horizontal range = u^2 sin2tita/g


then according to the condition....

max height =horizontal range

u^2 sin^2tita/2g = u^2 sin2tita/g

sin^2 tita/2 = sin2tita

sin^2tita/2 = 2 sintita costita

sin tita/2 = 2× costita

sintita/costita= 4

tan tita= 4

tita= (tan^-1)4

tita = tan^-1 (4)

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