Given, the maximum height reached by the car jumps from an inclined plane placed at one back of river is \(H_1=80 \mathrm{~m}\)
The second maximum height reached by car jumps from another inclined plane placed at other bank is \(\mathrm{H}_2=45 \mathrm{~m}\)
Let the certain velocity of moving car \(=u\)
Let the inclination angle in first jump \(=\boldsymbol{\theta}_1\)
Let the inclination angle in second jump \(=\theta_2\)
We know that,
the maximum height of projectile given as
\(H=\frac{u^2 \sin \theta}{2 g}\)
For first case,
\(\begin{aligned}
& H_1=\frac{u^2 \sin ^2 \theta_1}{2 g} \\
& 80=\frac{u^2}{2 g} \sin ^2 \theta_1 \quad \ldots (i)
\end{aligned}\)
Similarly, for second case, \(H_2=\frac{u^2}{2 g} \sin ^2 \theta_2\)...(ii)
Dividing Eqs. (i) by (ii), we get
\(\begin{aligned}
\therefore \quad \frac{H_1}{H_2} & =\frac{\frac{u^2}{2 g} \sin ^2 \theta_1}{\frac{u^2}{2 g} \sin ^2 \theta_2} \\
\Rightarrow \quad \frac{80}{45} & =\frac{\sin ^2 \theta_1}{\sin ^2 \theta_2} \\
\frac{\sin \theta_1}{\sin \theta_2} & =\frac{4}{3} \\
\Rightarrow \quad \frac{\sin \theta_1}{\sin \theta_2} & =\frac{\frac{4}{3}}{5}
\end{aligned}\)
Now, \(\sin \theta_1=\frac{4}{5}\) or \(\theta_1=\sin ^{-1}\left(\frac{4}{5}\right)=53^{\circ}\)
\(\therefore \quad \sin \theta_2=\frac{3}{5}\)
or \(\quad \theta_2=\sin ^{-1}\left(\frac{3}{5}\right)=37^{\circ}\)...(iv)
Now, from Eq. (i), we get
\(\begin{aligned}
& 80=\left(\frac{u^2}{g}\right)\left(\frac{\sin ^2 53}{2}\right) \text { or } 80=\left(\frac{u^2}{g}\right)\left[\frac{\left(\frac{4}{5}\right)^2}{2}\right] \\
& \frac{u^2}{g}=\frac{80 \times 25 \times 2}{16} \\
& \frac{u^2}{g}=250 \quad \ldots (v)
\end{aligned}\)
\(\therefore\) Width of the river given as
\(\begin{aligned}
& d=\left(\frac{u^2}{g}\right) \sin 2 \theta \\
& d=\frac{u^2}{g}\left(2 \sin \theta_1 \cos \theta_1\right) \\
& \quad(\because \sin 2 \theta=2 \sin \theta \cos \theta \text { and consider } \\
& \left.\qquad \theta=\theta_1 \text { for first case }\right)
\end{aligned}\)
From Eqs. (v) and (iii), we get
\(\begin{aligned}
& \therefore d=250 \times 2 \times \frac{4}{5} \times \frac{3}{5} \quad\left[\therefore \sin \theta_1=\frac{4}{5} \text { then } \cos \theta_1=\frac{3}{5}\right] \\
& d=20 \times 2 \times 4 \times 3=240 \mathrm{~m} \\
& d=240 \mathrm{~m}
\end{aligned}\)
Hence, the width of the river is \(240 \mathrm{~m}\).