Question

Calculate the projection angle of projectile whose range is $50m$ and maximum height $10m$

Answer 1

Given $R=50m;A=10m;\theta =?$
$\because RangeR=\frac{u^{2}sin2\theta }{g}$
And ,maximum height
$H=\frac{u^{2}si{n}^2\theta }{2g}$
$\therefore \frac{R}{H}=\frac{\frac{u^{2}sin2\theta }{g}}{\frac{u^{2}si{n}^2\theta }{2g}}=\frac{2sin2\theta }{si{n}^2\theta }=\frac{2\times 2sin\theta cos\theta }{si{n}^2\theta }$
or $\frac{50}{10}=4\frac{cos\theta }{sin\theta }=4cot\theta =\frac{4}{tan\theta }$
or $\frac{5}{1}=\frac{4}{tan\theta }=4cot\theta =\frac{4}{tan\theta }$
or $\frac{5}{1}=\frac{4}{tan\theta }\Rightarrow tan\theta =\frac{4}{5}=0.8$
$\therefore \theta =ta{n}^{-1}\left(0.8\right)={38.66}^0$
$\theta ={38.66}^0$