question_answer

    If a particle is thrown vertically upwards with a velocity of u cm/s under gravity, then the time for the particle to come to earth again is

A)  \[\frac{u}{g}s\]                               

B)  \[\frac{2u}{g}s\]

C)  \[\frac{u}{2g}s\]                             

D)  None of these

Correct Answer: B

Solution :

                Let the height of tower PQ is h and let point A is at a distance x from the foot\[p\]of the tower PQ. Now, ln \[\Delta APQ\]                 \[\tan \alpha =\frac{h}{x}\] \[\Rightarrow \]               \[h=x\tan \alpha \]                                    ...(i) In \[\Delta ABP,\]                 \[\tan \beta =\frac{b}{x}\] \[\Rightarrow \]               \[b=x\tan \beta \] \[\Rightarrow \]               \[x=\frac{b}{\tan \beta }\] Putting this value of\[x\]is Eq. (i), we get \[h=\frac{b}{\tan \beta }.\tan \alpha \] \[\Rightarrow \]      \[h=b\tan \alpha \cot \beta \]