question_answer

An aeroplane flying horizontally with a speed of \[360\,km{{h}^{-1}}\] releases a bomb at a height of 490 m from the ground. If \[g=9.8\,m{{s}^{-2}}\]. It will strike the ground at:

A)  10 km                  

B)         100 km

C)         1 km                    

D)         10 km

Correct Answer: C

Solution :

Horizontal speed of aeroplane \[\upsilon =360\,\text{kmph}\] \[=\frac{360\times 1000}{3600}\text{m}{{\text{s}}^{-1}}\] \[=100\,m/s\] Now,     \[s=ut+\frac{1}{2}g{{t}^{2}}\] \[\Rightarrow \]               \[490=\frac{1}{2}\times 9.8{{t}^{2}}\]                     \[[\therefore \,4=0]\] \[\Rightarrow \]               \[{{t}^{2}}=100\] \[\Rightarrow \]               \[t=10\,\sec \] Distance travelled along ground \[BC=\upsilon t\] \[=100\times 10\] \[=1000\,\,m\] \[=1\,km\]