Dealing with vertical motion quadratic equation?

Question

I need some physics help on where/why these numbers come out the way they do. I know that gravity is ~9.81 m/s/s and in feet that is about 32 ft/s/s. In math when we talk about vertical motion problems we use the equation d=rt-5t^2 for measuring in meters and we use d=rt-16t^2 for measuring in feet. MY QUESTION is WHY are we basically dividing by 2 from ~10 to 5 and ~32 to 16???

Answer 1

Because you are really solving a differential equation:

See the bottom of my answer for a non-calculus answer.

Apologies in advance if you haven't had any calculus yet, and are just working with the formulae without explanation of where they come from.

a = acceleration, v = velocity, h = distance, g = gravitational acceleration of the Earth, v0, h0 = initial velocity and height.
I use "h" instead of "d" to avoid confusion with the differentiation operator.

a = g = dv/dt => dv = gdt

integrate to get

v = gt +v0 = dh/dt => dh = gtdt +v0dt

Integrate to get

h = .5gt^2 + v0t + h0

Note the coefficient ".5g" for the t^2 term. This is due to the general form for integration of a polynomial - specifically the integral of t is 1/2t^2.

Non-calculus answer: (note that this only works for this specific case of constant acceleration).

Imagine you start at v =0. Drop a rock. How fast is it moving after 1 second? v1 = gt = 9.8 m/s

How far has it travelled in 1 second?
d = vt; but what is v? v was zero when we started and 9.8 after one second but it changed the entire time.

In this case we see that the velocity started out at 0 and ended at 9.8 m/s. So the average velocity is 1/2 of (0+9.8) = 4.9 m/s.

Using the average velocity we see the rock has dropped

d = vt = 4.9*1 = 4.9 meters.

generalizing we see that d = 1/2(0+at)t = 1/2at^2

Answer 2

~9.81 m/s/s is not correct, the units are m/sec^2 or m/s^2 using your notation. We say meter per second second (s^2), not meter per second per second (s/s).

Your equations d=rt-5t^2 and d=rt-16t^2 are not meaningful unless the variables are defined. What are d, r, and t? Even so, one can guess what they might be because the equations follow the pattern of one of the so-called SUVAT equations that govern the relationships between acceleration, distance, velocity, and time. [See source.]

The SUVAT equation you are apparently using is more commonly written as S = Vt - 1/2 At^2; where S = d, V = r, t = t and (1/2 A) = 5 and 16 because A = g = 9.81 m/sec^2 or 32.2 ft/sec^2. g is the acceleration due to gravity at Earth's surface. Thus, you can see that acceleration is divided by 2 in each equation.

The 1/2 in the SUVAT equation can be derived from more fundamental equations of motion by using calculus. But, without going into the calculus, suffice it to say that most of the time when there is a square (e.g., t^2) in a term, there will be a 1/2 in the term's coefficient. And that's why you divide the g = A by 1/2.

Answer 3

The equation is developed by stating that the acceleration due to gravity is -g. So -g = dv/dt. Integrating, you get v =-gt +c. Now ds/dt= -gt+c, so integrating we get s = -gt^/2+ct. Thus the equations you are using simply replace the value for g by whatever units you want and then divide it by 2 as required by the equation just developed. I used s for your d.

Answer 4

it has to do with calculus and the equations of motion where

acceleration (such as gravity) = dv/dt (rate of change in velocity with respect to time

and

Velocity = ds/dt (rate of change of distance with respect to time)

See link below for the fancy math part...

Answer 5

b) Make use of this equation: x = -4.9 t^2 + 15 t Find t (t>0) for which x = 0 (the ball is on the ground again)

Answer 6

s=ut+1/2*a*t^2
s=distance
u=initial velocity
t=time
a=acceleration
a]metric system
here a=-9.81m/sec^2
s=ut-4.905t^2
Dividing by two is beacause
of factor{1/2}*a*t^2

=ut-5t^2 approx.
b]fps system
here a=-32ft/sec^2
s=ut-16t^2