At time $t = 0$ a small ball is projected from point A with a velocity of $60{\text{m/s}}$ at $60^\circ $ angle with horizontal. Neglect atmospheric resistance and determine the two times ${t_1}{\text{ and }}{{\text{t}}_2}$ when the velocity of the ball makes an angle $45^\circ $ with the horizontal x-axis
(A) ${{\text{t}}_{\text{1}}} = 1.19s{\text{, }}{{\text{t}}_{\text{2}}} = 8.20s$
(B) ${{\text{t}}_{\text{1}}} = 2.19s{\text{, }}{{\text{t}}_{\text{2}}} = 7.20s$
(C) ${{\text{t}}_{\text{1}}} = 2.19s{\text{, }}{{\text{t}}_{\text{2}}} = 8.20s$

Hint This question will be a little difficult to understand. They provided the net velocity with which the ball travels. Resolve the two components of the velocity. When the velocity of the ball makes an angle $45^\circ $ the net time is the sum of ${t_1}{\text{ and }}{{\text{t}}_2}$. We have to find ${t_1}{\text{ and }}{{\text{t}}_2}$.

Complete step by step answer
Velocity: The velocity of an object is defined as the change of rate of its position with respect to time and direction i.e. velocity is the rate of change of displacement with respect to time.
There are three equations of motion. It is used to derive components such as displacement(s), velocity (initial and final), time (t) and acceleration (a).
First Equation of Motion \[ \to v{\text{ }} = {\text{ }}u{\text{ }} + {\text{ }}at\]
Second Equation of Motion \[ \to s{\text{ }} = {\text{ }}ut{\text{ }} + {\text{ }}\raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/
 \kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} {\text{ }}a{t^2}\]
Third Equation of Motion \[ \to {v^{2\;}} = {\text{ }}{u^2} + {\text{ }}2as\]
Where,
u is the initial velocity.
v is the final velocity.
a is the acceleration.
t is the time period .
s is the distance travelled i.e. displacement.
Given,
At time $t = 0$ a small ball is projected from point A velocity of $60{\text{m/s}}$ at $60^\circ $ angle with horizontal
The two times ${t_1}{\text{ and }}{{\text{t}}_2}$ when the velocity of the ball makes an angle $45^\circ $ with the horizontal x-axis will be?
We can find the velocities of the ball at time ${t_1}{\text{ and }}{{\text{t}}_2}$ using the first equation of motion.
First Equation of Motion \[ \to v{\text{ }} = {\text{ }}u{\text{ }} + {\text{ }}at\]
Which can be rewritten as
\[ \to v{\text{ }} = {\text{ }}u{\text{ }} - {\text{ }}gt\]
Here the acceleration is replaced with acceleration due to gravity and the minus symbol represents that the acceleration due to gravity is negative because all other acceleration will either be horizontal or upward but here the direction of acceleration due to gravity is downward.
At time $t = 0$ a small ball is projected from point A velocity of $60{\text{m/s}}$ at $60^\circ $ angle with horizontal
Let the velocity of the small ball at the time $t = 0$ be ${{\text{V}}_x}$
\[ \Rightarrow v{\text{ }} = {\text{ }}u{\text{ }} - {\text{ }}gt\]
\[ \Rightarrow {V_x}{\text{ }} = {\text{ 60}}\cos 60^\circ {\text{ }} - {\text{ 9}}{\text{.8}} \times 0\]
\[ \Rightarrow {V_x}{\text{ }} = {\text{ 60}}\cos 60^\circ {\text{ }} - {\text{ 0}}\]
\[ \Rightarrow {V_x}{\text{ }} = {\text{ 30 m/s}}\]
Let $t = {t_1} + {t_2}$
At times ${t_1}{\text{ and }}{{\text{t}}_2}$ when the velocity of the ball is
\[ \Rightarrow {V_y}{\text{ }} = {\text{ }}u{\text{ }} - {\text{ }}gt\]
\[ \Rightarrow {V_y}{\text{ }} = {\text{ 30sin30}}^\circ {\text{ }} - {\text{ 9}}{\text{.8}}t\]
\[ \Rightarrow {V_y}{\text{ }} = {\text{ 30}}\sqrt 3 {\text{ }} - {\text{ 9}}{\text{.8}}t\]
At times ${t_1}{\text{ and }}{{\text{t}}_2}$ the velocity of the ball makes an angle $45^\circ $ with the horizontal x-axis
$ \Rightarrow \tan \theta = \dfrac{{{V_x}}}{{{V_y}}}$
$ \Rightarrow \tan 45^\circ = \dfrac{{30}}{{30\sqrt 3 - 9.8t}}$
$ \Rightarrow \dfrac{{30\sqrt 3 - 9.8t}}{{30}} = \dfrac{1}{{\tan 45^\circ }}$
$ \Rightarrow \dfrac{{30\sqrt 3 }}{{30}} - \dfrac{{9.8t}}{{30}} = \dfrac{1}{{\tan 45^\circ }}$
$ \Rightarrow \sqrt 3 - 0.33t = \pm 1$
$ \Rightarrow 1.732 - 0.33t = \pm 1$
$ \Rightarrow t = \dfrac{{ \pm 1 + 1.732}}{{0.33}}$
$ \Rightarrow {t_1} = \dfrac{{ - 1 + 1.732}}{{0.33}}$
$ \Rightarrow {t_1} = 2.218s$
$ \Rightarrow {t_2} = \dfrac{{1 + 1.732}}{{0.33}}$
$ \Rightarrow {t_2} = 8.278s$
So, the time ${t_1}{\text{ and }}{{\text{t}}_2}$ is $2.218s{\text{ and }}8.278s{\text{ }}$

Hence the correct answer is option (C) ${{\text{t}}_{\text{1}}} = 2.19s{\text{, }}{{\text{t}}_{\text{2}}} = 8.20s$

Note We did not get the accurate value as the option C but we got the near approximation value of option C. Until the whole number is the same and the first two decimal numbers are nearer to the obtained answer it is fine.