The law of an object moving with constant acceleration is
Where 
is space, 
is time, 
is the initial position, 
is the initial velocity and 
is the acceleration.
In this case, if we choose a reference grid with the vertical axis pointing upwards, the acceleration of gravity will point downwards (and thus be negative). The initial position is zero, because the rocket is on the ground, and the initial velocity is 100 (positive because pointing upwards).
So, its law is
(I changed for 
since the rocket is moving vertically, so its position is actually its height. Also, g is the acceleration due to gravity).
The rocket hits the ground if its height is zero, so if we set 
we have
Solving for t, we have either t=0, or
The solution t=0 means that at the beginning the rocket is on the ground. So, we're interested in the other solution. Considering that g is about 32.2 feet/s^2, we have
Answer:
linear with a common first difference of 2
Step-by-step explanation:
If this is linear, the common first difference, or slope, will be the same throughout the entire table.
The formula for slope is

For the first set of points, we have:
m=(0--2)/(-2--3) = (0+2)/(-2+3) = 2/1 = 2
For the second set of points, we have:
m = (4-0)/(0--2) = 4/(0+2) = 4/2 = 2
For the third set of points, we have:
m = (12-4)/(4-0) = 8/4 = 2
The rate of change, or slope, is constant; this makes the function a linear function with a common first difference of 2
Hope it helps
Answer:
d≈56.01cm
A Area
2464
cm²
Using the formulas
A=πr2
d=2r
Solving ford
d=2A
π=2·2464
π≈56.01127cm
Step-by-step explanation:
Answer:
<h2>
a₁₈ = - 31</h2>
Step-by-step explanation:
