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I need help deriving the quadratic equation,

h= -16t^2+vt+s

where t is time, v is velocity, s is starting height, and h is the height of where the object will be.

I need to be able to explain where this formula comes from and why it is the way it is. I have been searching all over the net, but I can't find anything. Please help as soon as possible. Thanks.

2007-03-20 12:21:40 · 1 answers · asked by Micha 2 in Science & Mathematics Physics

1 answers

This equation is a function of the position of an object subjected to the force of gravity.

Start from Newton's Law:
F=m*a

examine the forces on an object in the vertical direction with respect to the Earth.
m*a=-m*g
or
a=-g

In the English units, g is modeled as a constant approximately 32 ft/sec ^2

the velocity of an object is the integral of the accleration in time.

Another way to think about this, is that the instantaneous rate of change of velocity of an object is the accleration of an object. Or, expressed differentially:

That is dv(t)/dt = a(t)

IN this specific case, since the gravity is approximately constant,
dv(t)/dt=-g

so the rate of change of velocity for vertical motion of a mass acted on by gravity is -g

integrate to find that v(t)=-g*t+C
where C is a constant.

To determine C, evaluate the motion of the object at time zero, t=0

If the object has velocity in the vertical direction at t=0, we call that v0, which you have as simply v in your equation.

so now we know that
v(t)=-g*t+v

position is known to be the integral of velocity in time.

That is to say, the instantaneous rate of change of the position of an object is the velocity of that object.
Or, expressed differentially:
dh(t)/dt=v(t)

since we know that v(t)=-g*t+v

we can integrate to find that
h(t)=-.5*g*t^2+v*t+C

again, we have to evaluate the function at t=0 to determine the constant C. This is the position of the object at time equals zero. You have this as s in your equation, which is the known starting position of the object.

so
h(t)=-.5*g*t^2+v*t+s

plugging in the English units for g
h(t)=-16*t^2+v*t+s

=============================
Here's another derivation without assuming knowledge of calculus.
This requires an important assumption: that the gravitational force is a constant. This is approximately true and approximately 32 ft/sec^2 for motion much less than the speed of light and relatively small change in position with respect to the surface of the Earth. Relatively small can be a few miles from the surface.

Starting from Newton's first Law, the law of inertia, an object that is not acted upon by an external force will not experience a change in velocity.

If we observe an object having velocity in the vertical direction at a time we'll say is t=0 for this observation, and that velocity is v, then that object will continue to have that velocity unless acted upon by an external force.

We will also set our sign convention as positive in the upward directin, which we assume to be radially away from the center of the Earth.

If this object is in the proximity of the surface of the Earth, we know through experimental evidence that there is a force due to gravity which is also radial with respect to the surface of the Earth. According to Newton's second Law, the rate of change of velocity of an object is proportional to the force acting on the object. Again, we determine that the acceleration is a constant, and downward, or -g.

Combining the first and second laws together, we deduce that
v(t)=-g*t+v

If we observe an object having position, s, at time equals zero, that is an unchangeable fact.

therefore,
h(0)=s

we also know that the change in position of an object is proportional to the velocity of that object.

If the velocity is constant, then the position is
h(t)=s+v*t, since the object will change position at a constant rate.

Bring in a force,gravity in this case, and the velocity is variable with respect to time. Through observation, the position is
h(t)=s+v*t-.5*g*t^2
or
h(t)=-16*t^2+v*t+s

j

2007-03-20 12:33:17 · answer #1 · answered by odu83 7 · 1 0

Vertical Motion Formula

2016-09-30 01:59:16 · answer #2 · answered by ? 3 · 0 0

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RE:
Vertical Motion Formula Help! Algebra Derivation Needed!?
I need help deriving the quadratic equation,

h= -16t^2+vt+s

where t is time, v is velocity, s is starting height, and h is the height of where the object will be.

I need to be able to explain where this formula comes from and why it is the way it is. I have been searching all over the...

2015-08-17 02:04:02 · answer #3 · answered by Anonymous · 0 0

set up your questions: v0= 250 ft/s y0= 100 ft t0= 0 sec t= 3 sec g= 32.2 ft/s^2

2016-03-14 13:46:16 · answer #4 · answered by Anonymous · 0 0

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