The ratio of maximum range to maximum height of a projectile is 4. Determine the angle of projection.
A.
\(30^0\)
B.
\(60^0\)
C.
\(45^0\)
D.
\(0^0\)
Correct answer is C
Maximum Range R = \(\frac{u^2sin2\theta}{g}\)
maximum height H = \(\frac{u^2sin^2\theta}{2g}\)
But R : H = 4
\(\frac{u^2sin2\theta}{g}\) : \(\frac{u^2sin^2\theta}{2g}\) = 4
\(\frac{u^2sin2\theta}{g} \times\frac{2g}{u^2sin^2\theta}\) = 4
\(\frac{2sin2\theta}{sin^2\theta}\) = 4
since 2sinΘ = 2sinΘcosΘ
\(\frac{2 \times 2sin\theta cos\theta}{sin^2\theta}\) = 4
\(\frac{4sin\theta cos\theta}{sin^2\theta}\) = 4
\(\frac{sin\theta cos\theta}{sin^2\theta}\) = 1
\(\frac{cos\theta}{sin\theta}\) = 1
\(\frac{sin\theta}{cos\theta}\) = 1
\(tan \theta\) = 1
\(\theta = tan^{-1}1\) = 45°
