NEET Sample Paper NEET Sample Test Paper-1

  • question_answer A bomber plane is moving horizontally with a speed of \[500m/s\] and bomb released from it, strikes the ground in 10 seconds. Angle at which it strikes the ground is \[(g=10m/{{s}^{2}})\]:

    A)  \[{{\sin }^{-1}}\frac{1}{5}\]                     

    B)  \[ta{{n}^{-1}}\frac{1}{5}\]

    C)  \[ta{{n}^{-1}}\,5\]              

    D)  \[ta{{n}^{-1}}\,1\]

    Correct Answer: B

    Solution :

    Here: Speed of the bomber plane \[V=\text{ }500\text{ }m/s\]. Time taken by the bomb to strike the ground \[=10\sec .\]                         \[g=10\,m/{{s}^{2}}\] \[\therefore \]   Time is given by             \[t=\sqrt{\frac{2h}{g}}\] \[\Rightarrow \]               \[10=\sqrt{\frac{2h}{g}}\] \[\Rightarrow \]               \[\frac{2h}{g}={{(10)}^{2}}=100\] \[\Rightarrow \]               \[2h=100\times 10\] \[\Rightarrow \]               \[h=500\,m.\] and vertical velocity is given by             \[V=\sqrt{2gh}\]                 \[V=\sqrt{2\times 10\times 500}=100\,m/s.\] Therefore,     \[\tan \theta =\frac{vertical\text{ }velocity}{horizontal\text{ }velocity}\]                 \[=\frac{100}{500}=\frac{1}{5}\]                 \[\theta ={{\tan }^{-1}}\left( \frac{1}{5} \right)\]


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