Question

http://prntscr.com/b16ajw Help please?

Answer 1

Note that where gravity is involved, and the motion vertical, the applicable equation of motion is \[at^2+v _{0}t+s _{0}=s,\]

Answer 2

and in this case, assuming that we use the English system of measurement, the acceleration, a, would be \[-32\frac{ ft }{ \sec^2 }\]

Answer 3

and the initial velocity 12 m/sec (metric system). Ignore my -32 (above) and substitute -9.8 m / (sec^2) (metric system).

Hope this helps.

Answer 4

Okay. Thank you Mathmale!
The answer should be C. \[-4.9t^2+12t=2\] Right?

Answer 5

Could you explain why you believe that '2' should be on the right side of the equation?
That "2' represents the initial height of the ball off the ground.