Trajectories of two particles projected from origin with speed $v_1$ and $v_2$ and angles ${\theta }_1$ and ${\theta }_2$ with positive $x-$ axis, respectively, are as shown in the figure. Given that $g=10m/s^2$, find the ratio of ${\theta }_1$ to ${\theta }_2$.

Step 1: Find the angle ${\theta }_1$.

Formula used: $4H=R\tan \theta$

Given, maximum height for first particle, $H_1=10m$

And range of the particle $1,R_1=40m$

Using the relation, $4H_1=R_1\tan {\theta }_1$

$4\times 10=40\left(\tan {\theta }_1\right)$

$\Rightarrow {\theta }_1=\frac{\pi }{4}$

Step 2: Find the angle ${\theta }_2$

Formula used: $4H=R\tan \theta$

Given, maximum height for second particle,
$H_2=15m$

And range of the particle $2,R_2=20\sqrt{3}m$

Using the relation, $4H_2=R_2\tan {\theta }_2$

$4\times 15=20\sqrt{3}\left(\tan {\theta }_2\right)$

$\Rightarrow {\theta }_2=\frac{\pi }{3}$

Step 3: Find the ratio of ${\theta }_{1}to{\theta }_2.$

The ratio of ${\theta }_{1}to{\theta }_2.$

$\frac{{\theta }_1}{{\theta }_2}=\frac{\pi }{4}\times \frac{3}{\pi }=0.75$

Final answer: 0.75