A projectile is fired at an angle of $45{^\circ}$ with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection, is:

Question

Dr. Abubacker · Accepted Answer

(d) $H = \frac{u^{2}\sin^{2}{45{^\circ}}}{2g} = \frac{u^{2}}{4g}$ …(i)
$R = \frac{u^{2}\sin{90{^\circ}}}{g} = \frac{u^{2}}{g}$
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$	herefore\frac{R}{2} = \frac{u^{2}}{2g}$ …(ii)
$	herefore	an\alpha = \frac{H}{R/2}$
$= \frac{\frac{u^{2}}{4g}}{\frac{u^{2}}{2g}} = \frac{1}{2}\ 	herefore\ \alpha = 	an^{- 1}\left( \frac{1}{2} ight)$

Q. A projectile is fired at an angle of $45{^\circ}$ with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection, is:

Explanation

Reviewed by Dr. Abubacker

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