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(D) Given, the maximum height reached by the car in the first jump is $H_1 = 80 \,m$.The maximum height reached by the car in the second jump is $H_2

Vedclass · Accepted Answer

$240$

Vedclass · Answer

(D) Given, the maximum height reached by the car in the first jump is $H_1 = 80 \,m$.The maximum height reached by the car in the second jump is $H_2 = 45 \,m$.Let the initial velocity of the car be $u$ and the angles of inclination be $	heta_1$ and $	heta_2$ respectively.The maximum height of a projectile is given by $H = \frac{u^2 \sin^2 	heta}{2g}$.For the first case: $H_1 = \frac{u^2 \sin^2 	heta_1}{2g} = 80 \,m$ ... $(i)$For the second case: $H_2 = \frac{u^2 \sin^2 	heta_2}{2g} = 45 \,m$ ... (ii)Dividing $(i)$ by (ii): $\frac{H_1}{H_2} = \frac{\sin^2 	heta_1}{\sin^2 	heta_2} = \frac{80}{45} = \frac{16}{9}$.Taking the square root: $\frac{\sin 	heta_1}{\sin 	heta_2} = \frac{4}{3}$.Since the car reaches the same point on the other bank, the horizontal range $R$ is the same for both jumps.$R = \frac{u^2 \sin 2	heta}{g} = \frac{2u^2 \sin 	heta \cos 	heta}{g}$.Since $R_1 = R_2$, we have $\sin 	heta_1 \cos 	heta_1 = \sin 	heta_2 \cos 	heta_2$, which implies $\sin 2	heta_1 = \sin 2	heta_2$. This occurs if $	heta_2 = 90^\circ - 	heta_1$, meaning $\cos 	heta_1 = \sin 	heta_2$.From $\frac{\sin 	heta_1}{\sin 	heta_2} = \frac{4}{3}$, we get $\frac{\sin 	heta_1}{\cos 	heta_1} = 	an 	heta_1 = \frac{4}{3}$.Thus, $\sin 	heta_1 = \frac{4}{5}$ and $\cos 	heta_1 = \frac{3}{5}$.From $(i)$, $\frac{u^2}{2g} (\frac{4}{5})^2 = 80 \Rightarrow \frac{u^2}{g} = 80 	imes 2 	imes \frac{25}{16} = 250$.The range $d = \frac{u^2 \sin 2	heta_1}{g} = \frac{u^2}{g} (2 \sin 	heta_1 \cos 	heta_1) = 250 	imes 2 	imes \frac{4}{5} 	imes \frac{3}{5} = 240 \,m$.

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