The ratio of maximum range to maximum height of a projectile is 4. Determine the…

Explanation

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°