A ball is suspended by a thread from the ceiling of a car. The brakes are applied and the speed of the car changes uniformly from $\mathop {36kmh}\nolimits^{ - 1} $ to zero in $5s$ . The angle by which the ball deviates from the vertical (\[g = 10m{s^{ - 2}}\]) is:
(A) \[ta{n^{ - 1}}(\frac{1}{3})\]
(B) \[si{n^{ - 1}}(\frac{1}{5})\]
(C) \[ta{n^{ - 1}}(\frac{1}{5})\]
(D) \[co{t^{ - 1}}(\frac{1}{3})\]

Hint We will first find out the acceleration of the car. Finally we will find out the angle it makes with the vertical.
Formulae Used: $a = \frac{{(v - u)}}{t}$
Where, $a$ is the acceleration of the body, $v$ is the final velocity, $u$ is the initial velocity and $t$ is the time taken by the body.

Complete Step By Step Solution
Here,
\[u = {\text{ }}36{\text{ }}km{h^{ - 1}} = 10m{s^{ - 1}}\]
\[v = 0km{h^{ - 1}} = 0m{s^{ - 1}}\]
\[t = 5s\]
Now,
\[{a_{car}} = \frac{{(0 - 10)}}{5}\]
Now,
\[{a_{}} = - 2m{s^{ - 2}}\]
\[\begin{gathered}
  tan\theta = \frac{a}{g} = \frac{2}{{10}} = \frac{1}{5} \\
    \\
\end{gathered} \]
Thus, the answer turns out to be \[\theta = ta{n^{ - 1}}(\frac{1}{5})\] which is (C).
But,
Also,
If we take
\[\begin{gathered}
  sin\theta = \frac{2}{{10}} = \frac{1}{5} \\
    \\
\end{gathered} \]
\[ \Rightarrow \theta = si{n^{ - 1}}(\frac{1}{5})\]

Hence, (B) is also correct.

Note We got the value of \[sin\theta \] , we took \[{a_{net}} = \sqrt {{a^2} + {g^2}} \] an d value turns out to be \[10m{s^{ - 2}}\]. Also, we took only the magnitude as it only plays the role to form the angle.