A car moving with a certain velocity, jumps from an inclined plane placed at one bank of a river and reaches…
A car moving with a certain velocity, jumps from an inclined plane placed at one bank of a river and reaches the other bank by attained a maximum height of \(80 \mathrm{~m}\). If the same car, moving with the same velocity jumps from another inclined plane having different angle of inclination and reaches the same point on the other bank by attaining maximum height of \(45 \mathrm{~m}\), then the width of the river is
\(80 \mathrm{~m}\)
\(60 \mathrm{~m}\)
\(125 \mathrm{~m}\)
\(240 \mathrm{~m}\)
Solution
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 \(240 \mathrm{~m}\).