Given data
*The value of the acceleration due to gravity is a = 10 m/s^2
*The given time is t = 6 s
*The initial speed of the rock is u = 0 m/s
The relation for the velocity, acceleration, and time is given by the kinematic equation of motion as
[tex]v=u+at[/tex]
Substitute the known values in the above expression as
[tex]\begin{gathered} v=0+(10)(6)_{} \\ =60\text{ m/s} \end{gathered}[/tex]
The speed of the rock in MPH is converted as
[tex]\begin{gathered} v=(60\text{ m/s) \times}\frac{2.2369\text{ mph}}{1\frac{m}{s}} \\ =134.21\text{ mph} \end{gathered}[/tex]
Hence, the speed of the rock is v = 134.21 mph
