Answer:
(a) [tex]v_{f}=26.08m/s[/tex]
(b) [tex]t=4.906seconds[/tex]
Explanation:
We will take ground level as origin and upward is positive direction
the givens are
[tex]y_{i}=0\\y_{f}=-9.0m\\ v_{i}=22m/s[/tex]
Part (a)
To find the final velocity we use the kinematic equation
So
[tex](v_{f})^{2} =(v_{i})^{2} +2a(y_{f}-y_{i})\\(v_{f})^{2}=(22m/s)^{2}+2(-9.8m/s^{2} ) (-10-0)\\v_{f}=\sqrt{(22m/s)^{2}+2(-9.8m/s^{2} ) (-10-0)} \\v_{f}=26.08m/s[/tex]
Part (b)
To find time of rock trip until it touches the ground we will use simple kinematic equation or simple motion equation
[tex]v_{f}=v_{i}+at\\t=\frac{v_{f}-v_{i}}{a}\\ t=\frac{(-26.08m/s)-(22m/s)}{-9.8m/s^{2} }\\t=4.906seconds[/tex]
Notice that we substituted vf with negative sign because its direction is downwards
