UPSEE Solved Model Paper 7

Given ： Nozzle velocity of the defense cannon, u = 85 m/s Distance between the defense cannon and the enemy ship, d = 575 m The cannon will execute a projectile motion and hit the enemy ship Thus, Distance between the defense cannon and the enemy ship = Horizontal range of the cannon We know that the horizontal range is given by R = $\frac{u^2 \sin(2\theta)}{g}$ ... (i) ⇒ sin (2θ) = $\frac{Rg}{u^2}$ ⇒ θ = $\frac{1}{2} \times \sin^{-1}\left(\frac{Rg}{u^2}\right)$ ⇒ θ = $\frac{1}{2} \times \sin^{-1} \left[ \frac{575\,\text{m} \times 9.8\,\text{m/s}^2}{(85\,\text{m/s})^2} \right]$ ⇒ θ = $\frac{1}{2} \times \sin^{-1}(0.78)$ Using $\sin^{-1}$ 0.78 = 51.26° , we get θ = $\frac{1}{2}$ × 51.26° ⇒ θ = 25.63° Now, re-writing eq (i) as R = $\frac{u^2 \sin(2\theta)}{g}$ = $\frac{2u^2 \sin\theta \cos\theta}{g}$ [as sin (2θ) = 2 sin θ cos θ] If we change θ to "90° - θ" we have R' = $\frac{2u^2 \sin(90^{\circ} - \theta) \cos(90^{\circ} - \theta)}{g}$ ⇒ R' = $\frac{2u^2 \cos\theta \sin\theta}{g}$ = $\frac{u^2 \sin(2\theta)}{g}$ = R i.e., range remains the same Therefore, the corresponding angles for the cannon are 25.63° and 90° - 25.63° = 64.37° The commandant of the fort can elevate the gun to either if the two angles and hit the enemy ship Hence, option 'C' is correct