Question

Find the angle of projection at which horizontal range and maximum height are equal.
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Answer 1

Answer:

The angle of projection,

Horizontal range and maximum height are equal

⇒ For horizontal range,

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

⇒ For maximum height,

$R= \frac{u^{2} sin2 \theta}{2g}$

⇒ so, H=R

$\frac{u^{2} sin^{2}\theta }{2g} = \frac{u^{2} sin{2}\theta }{g}$

⇒  then,  $\frac{sin^{2} \theta}{2} =sin2\theta$

$sin^{2} \theta =2 \ or \ tan\theta=4$

so, $\theta=tan^{-1} (4)$

so the theta is tan-1(4).