Answer :
Given that the rock is moving upward, so it will have an acceleration due to gravity, g = -9.81 m/s^2
When time, t =2 seconds, the velocity is v = 13m/s.
We have to find (a) speed at the time of launch
(b) speed of rock when time , t' =4.6 s
(a) Let the launch speed be u.
Also, the displacement of the rock after 2 s will be
[tex]\begin{gathered} s=v\times t \\ =13\times2\text{ m} \\ =26\text{ m} \end{gathered}[/tex]Using the equation of motion,
[tex]s=ut-\frac{1}{2}gt^2[/tex]The launch speed can be calculated as
[tex]u=\frac{s+\frac{1}{2}gt^2}{t}[/tex]Substituting the values, speed will be
[tex]\begin{gathered} u=\frac{26+\frac{1}{2}\times9.81\times(2)^2}{2} \\ =22.81\text{ m/s} \end{gathered}[/tex]Thus, the launch speed is 22.81 m/s
(b) The time is t'=4.6 s.
Let the speed of the rock at this time be v'.
The speed can be calculated by the formula,
[tex]v^{\prime}=u-gt[/tex]Substituting the values, the speed will be
[tex]\begin{gathered} v^{\prime}=22.81-9.81\times4.6 \\ =-22.316\text{ m/s} \end{gathered}[/tex]Here, the negative sign of speed indicates that the rock is moving downwards at this time.
The magnitude of speed at t'=4.6 s is 22.316 m/s