UPSEE Solved Model Paper 7

The schematic diagram is shown below: Given: Horizontal velocity of the aeroplane, u = 150 m/s Height of the aeroplane, h = 80 m As the bomb is just dropped, its initial vertical velocity is zero Consider Y-motion of the bomb. Using second equation of motion: S = ut + $\frac{1}{2} a t^{2}$ , we have S = h a = g (acceleration due to gravity) u = 0 Thus, h = $(0 \times t) + \frac{1}{2} g t^{2}$ ⇒ t = $\sqrt{\frac{2h}{g}}$ ⇒ t = $\sqrt{\frac{2 \times 80 \mathrm{m}}{10 \mathrm{m/s}^2}}$ = 4s This is the time taken by the bomb to reach the ground (or at point B, target). [Note: Time is a scalar quantity. So it does not matter which motion we consider to calculate it.] Now, the initial horizontal velocity of the bomb being dropped is the same as that of the aeroplane. Also, as there is no horizontal force acting on the bomb, its horizontal velocity remains constant (neglecting friction). Thus, the distance covered by the bomb is AB = u × t ⇒ A B = 150 m/s × 4 s ⇒ AB = 600 m Using Pythagoras theorem in the right Δ ABC, the distance between the dropping point and the target is CB = $\sqrt{AC^{2}+AB^{2}}$ ⇒ CB = $\sqrt{(80\mathrm{m})^{2}+(600\mathrm{m})^{2}}$ ⇒ CB = 605.3 m Hence, option 'A' is correct