A=g
v=⌠g dt
v=gt+C, where C=v initial
v=gt+vi
h=⌠v dt
h=gt^2/2+vit+C, where C=h initial
h=gt^2/2+vit+hi
We are told that vi=42 ft/s, hi=0, and we know g≈-32 ft/s^2 so
h(t)=-16t^2+42t
The ball will hit the ground when h=0 so
-16t^2+42t=0
-2t(8t-21)=0, since t>0
8t-21=0
8t=21
t=21/8
t=2.625 sec
t≈2.6 sec (to nearest tenth of a second)
<u>The answer is 808/99</u>
Answer:
it goes 0 then 1/4 in the first notch 1/2 in the second notch 3/4 in the third notch then the number 1 on the fourth
Step-by-step explanation:
Hope that helps.
Answer:
y = - 8x² + 6
Step-by-step explanation:
The equation of a parabola in vertex form is
y = a(x - h)² + k
where (h, k) are the coordinates of the vertex and a is a multiplier
Here (h, k) = (0, 6), thus
y = a(x - 0)² + 6, that is
y = ax² + 6
To find a substitute (- 1, - 2) into the equation
- 2 = a(- 1)² + 6, that is
- 2 = a + 6 ( subtract 6 from both sides )
- 8 = a
y = - 8x² + 6
