Actually the position function with respect to time under constant acceleration is:
a=g
v=⌠g dt
v=gt+vi
s=⌠v
s=gt^2/2+vit+si
So if vi and si are zero then you just have:
s=gt^2/2
Notice that it is not gt^2 but (g/2) t^2
So the first term in any quadratic is half of the acceleration times time squared because of how the integration works out...
Anyway....
sf=(a/2)t^2+vit+si
(sf-si)-vit=a(t^2)/2
2(sf-si)-2vit=at^2
a=(2(sf-si)-2vit)/t^2 and if si and vi equal zero
a=(2s)/t^2
Answer:
Step-by-step explanation: base x height
70ft
Answer:
<em>(B). 319 </em>
Step-by-step explanation:
= 
=
x = 
≈ <em>319</em>
