Question

http://prntscr.com/b18f2m Correct?

Answer 1

Ok so this is a one direction motion, with gravity being the force.

Let's go through our diff. eqns

\[y'' = -g\]
\[y' = -g t + C \] when t = 0, y' = 12 (initial velocity of 12), so C = 12
\[y = \int\limits (-g t +12) dt = -\frac{ 1 }{ 2 }g t^2 +12t + C\]
When t = 0, y = 2 (initially 2 m above grounds)
So y = 2 = 0+ 0 + C
\[y = 12t -\frac{ 1}{ 2 } g t^2 +2\]

Answer 2

The last part is make your equation for height y = 0 if the ball is on the ground
so 12t - 0.49gt^2 + 2 = 0

Answer 3

sorry should say 12t - 4.9t^2 + 2 = 0 which is second response.

Answer 4

This can also be done using the formula\[d=v _{o}t+\frac{ 1 }{ 2 }g t ^{2}\]

Answer 5

Because the ball starts two meters above the ground, the ball on the ground is a negative direction, or a displacement of -2 meters.