A) 15 cm above the target
B) 10 cm above the target
C) 2.2 cm above the target
D) directly towards the target
Correct Answer: C
Solution :
| [c] The bullet performs a horizontal journey of 100 cm with constant velocity of 1500 m/s. The bullet also performs a vertical journey of h with zero initial velocity and downward acceleration g. |
| \[\therefore \] For horizontal journey, \[\text{time }\left( \text{t} \right)=\frac{\text{Distance}}{\text{Velocity}}\] |
| \[\therefore \,\,\,t=\frac{100}{1500}=\frac{1}{15}\text{sec }...\text{(1)}\] |
| The bullet performs vertical journey for this time. |
| For vertical journey, \[\text{h}\,\text{=}\,\text{ut+}\frac{\text{1}}{\text{2}}\text{g}{{\text{t}}^{\text{2}}}\] |
| \[\text{h=0+}\frac{1}{2}\times 10\times {{\left( \frac{1}{15} \right)}^{2}}\] |
| \[\text{or, h=}\frac{20}{9}\text{ cm = 2}\text{.2cm}\] |
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