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:
15 ft
Step-by-step explanation:
Hi, the illustration for the problem is a right triangle with:
hypotenuse (C)= the length of the ladder = L
horizontal side(A) = distance from bottom of the ladder to the building = L - 6
vertical side(B) = distance from the top of the ladder to the bottom of the building = L - 3
So, we can use Pythagoras formula:
A2 +B2= C2
(L – 6 )² + (L-3)² = L²
L²-12L+36+L²-6L+9 =L²
L2 -18L+45 =0
APPLYING QUADRATIC FORMULA WE OBTAIN:
L =15 OR L=3
If L=3
Vertical side = L-3 = 0 (Length can´t be 0)
So L=15
Answer:
x>4
Step-by-step explanation:
