Question

If $R$ and $H$ represent horizontal range and maximum height of the projectile, then the angle of projection with the horizontal is:

• A. $ta{n}^{-1}\left(H/R\right)$
• B. $ta{n}^{-1}\left(2H/R\right)$
• C. $ta{n}^{-1}\left(4H/R\right)$
• D. $ta{n}^{-1}\left(4R/H\right)$

Answer 1

Answer: (C)

$ta{n}^{-1}\left(4H/R\right)$

Maximum height, H = $\frac{u^{2}si{n}^2\theta }{2g}$-----(i)
Horizontal range, R = $\frac{u^{2}sin2\theta }{g}$ = $\frac{2u^{2}sin\theta cos\theta }{g}$------(ii)
Divide (i) by (ii), we get
$\frac{H}{R}=\frac{tan\theta }{4}ortan\theta =4H/Ror\theta =ta{n}^{-1}\left(4H/R\right)$