A) 51 sec
B) 25.5 sec
C) 61 sec
D) 21 sec
Correct Answer: A
Solution :
Here: Velocity of bomb \[\upsilon =1000\text{ }m/s\] The angle \[\theta =30{}^\circ \] Time taken by the bomb to reach the highest point \[t=\frac{\upsilon \sin \theta }{g}\] \[=\frac{1000\sin 30{}^\circ }{9.8}\] So, \[t=\frac{1000\times 0.5}{9.8}=51\,\sec \]You need to login to perform this action.
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