A) 605.3 m
B) 600 m
C) 80 m
D) 230 m
Correct Answer: B
Solution :
Time taken by the bomb to reach the ground is given by \[{{h}_{{{O}_{A}}}}=\frac{1}{2}gt_{OB}^{2}\] we have \[{{t}_{OB}}=\sqrt{\frac{2{{h}_{OA}}}{g}}\] Given, \[{{h}_{OB}}=80\,\,m,g=10\,\,m/{{s}^{2}}\] \[\therefore \] \[{{t}_{OB}}=\sqrt{\frac{2\times 80}{10}}=4\,s\] Horizontal velocity of bomb v = 150 m/s Horizontal distance covered by the bomb \[AB=v{{t}_{OB}}\] \[150\times 4\] = 600 m Hence, the bomb should be dropped 600 m before the target.You need to login to perform this action.
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