Answer:
(a) [tex]v_0= 34m/s[/tex]
(b) [tex]v=14m/s[/tex]
Explanation:
We can derive the initial speed of the rock from the equation of the speed in function of the time:
[tex]v=v_0-gt\\\\\implies v_0=v+gt[/tex]
Using the given values for the speed at time t=1.7s, we get:
[tex]v_0=17m/s+(9.8m/s^{2})(1.7s)=34m/s[/tex]
In words, the speed of the rock at launch is 34m/s (a).
Next, we use this to calculate the speed at t=4.9s:
[tex]v=v_0-gt\\\\v= 34m/s-(9.8m/s^{2})(4.9s)=-14m/s[/tex]
This means that the speed of the rock at 4.9s after the launch is 14m/s (b), and the negative sign means that it is moving downwards.
