Question

A particle takes \[\left( {7 + 2\sqrt 6 } \right)\]s and \[\left( {7 - 2\sqrt 6 } \right)\]s to reach foot of the building when projected vertically upward and downward respectively with same speed from the top of the building. The time taken by the particle to reach foot of the building when it is dropped freely from the top of the building is:
a. \[10s\]
b. \[5s\]
c. \[8s\]
d. \[4s\]

Answer 1

Hint: Here, the concept we are going to use is of vertical motion whether it is upward or downward. But, here both are mentioned here and the speed here is same. We have to calculate the total time taken by the particle.

Formula used:
\[H = ut + \dfrac{1}{2}g{t^2}\], which is nothing but the second kinematical equation.
Where, \[H\]stands for height (displacement), \[u\]stands for speed, \[g\] for acceleration and \[t\] stands for time.

Complete answer:
Let us solve this question, according to given data we have
\[{t_1} = \left( {7 + 2\sqrt 6 } \right)\] and \[{t_2} = \left( {7 - 2\sqrt 6 } \right)\]

We have the formula for solving this question as mentioned in the formula we used.
\[H = ut + \dfrac{1}{2}g{t^2}\]

Now, by rearranging the above formula we get
\[{t^2} + \dfrac{{2ut}}{g} - \dfrac{{2H}}{g} = 0\]

This formula looks like a quadratic equation and we have time as the roots of this equation i.e,. \[({t_1}\times {t_2})\].

By using product of roots in above equation, we get
${t_1}{t_2}=\left( {7 + 2\sqrt 6 } \right) \times \left( {7 - 2\sqrt 6 } \right)$
${t_1}{t_2}=49-\left(4\times6\right)=49-24=25$

By putting this value, we get
\[ \Rightarrow 25 = \dfrac{{ - 2H}}{g}\] …
\[ \Rightarrow H = - 125\]

But height can never be negative so the height is \[125\] and negative sign shows downward travel of the particle. Now, time taken to fall after release is given by:
\[T = \sqrt {\dfrac{{2H}}{g}} = \sqrt {\dfrac{{2 \times 125}}{{10}}} = 5s\]
\[\therefore T = 5s\]

Thus, the time taken is \[5s\]. Correct option is b.

Note: For more information we have to know that the magnetic field is equal to the horizontal compass’s magnetic field though there is deflection. The magnetic moment is calculated using the formula for magnetic field of the magnet..