# Relation between the gravitional constant and the given equation

1. Jun 14, 2007

### threetheoreom

1. The problem statement, all variables and given/known data

The problem says the general formula for the height (h) after (t)seconds of a projectile launched upward from ground level with initial speed (v) is h = vt-1/2gt^2, where (g) is the gravitational constant (appox. 32m/s^2).

Show That the greatest height of the projectile is v^2/2g

2. Relevant equations

3. The attempt at a solution

I tried alot a stuff .... I think The problem I am having here is making a relation between the gravitional constant and the given equation.

Thanks guys.

Last edited: Jun 14, 2007
2. Jun 14, 2007

### threetheoreom

The is probably not even close but i tried this: I made the the greatest height rational expression into a f(0), that allows me to think that the higher the function goes closer it gets to zero, and this depends on time (t) seconds, so I relized that all i have to do is make the equation = 0.

So h-vt-1/2gt^2=0 . ( is that correct ) but now thw question is how do relate the gravitational constant. Now constant gives me a since polynomial quadratic equations are in the form ax^2 + bx + cx =0 Then maybe I should try plugging the s^2 into the equ. so h-vs^2 + 1/2gt^2 should be it but i cant solve for that. Where does my resoning blurr?

3. Jun 14, 2007

### Ks. Jan Jenkins

Firstly, some basic equations of motion, which one can derive using integral calculus:

a = -g (the acceleration is in the opposite direction of the initial velocity, that is, downwards)

v = v(0) + a*t (the velocity at any given moment is equal to the initial velocity plus the acceleration multiplied by the time it has accelerated).

The equation given actually uses v for the variable v(0) here. Acceleration is -g, thus one has:

velocity = v - g * t

For the position of the projectile at time t:

s = 1/2 a * t^2 + v(0) * t + s(0)

which using the variable names in your equation :

h = v * t - 1/2 g * t^2 + 0 (s(0) is 0 as the problem says that the projectile is launched from ground level = 0)

This equation for h (s for position or 'spatium' in latin) gives the height of the projectile at any certain time t.

Thus rephrasing your question: what is the greatest height of the projectile?

This is equivalent to saying: what is the maximum of this function? You should know that the minimum or maximum of a function always occurs when the (first) derivative of the function = 0. So let us take the first derivative of s with respect to t:

ds/dt = v - 1/2 g * 2t + 0

thus t = v / g at the maximum of this function. Substitute t into your original equation for s and you will find out what s must be at that time t.

s = v * v / g - 1/2 g * (v/g)^2
s = v^2/g - 1/2 g * v^2 / g^2
s = (v^2 - 1/2 v^2)/g
s= 1/2 v^2/g or (v^2)/(2*g)

Which is the answer to your problem - the height is = 1/2 v^2/g at it's maximum.

Bonus question:

The project is launched at an angle A from the horizontal. What is the range of the projectile with the initial velocity v? What is its maximum height? At what angle will the range be the maximum?

4. Jun 14, 2007

### Ks. Jan Jenkins

Oops - I just noticed that this forum is for precalculus mathematics - sorry for the answer a little too far ahead.

However, as you noted before you see that the velocity gets smaller and smaller - just find out when it equals zero (it will equal zero when it reaches its maximum height) and then substitute that time into the height equation.

JJ +

Last edited: Jun 14, 2007
5. Jun 14, 2007

### threetheoreom

Thanks this solution let me understand and finish my own argument and logic ..even though the math is a little beyond me.

6. Jun 17, 2007

### Feldoh

I think I figured out the range...
$$d_x = v_{ox}t$$
$$d_y = v_{oy}t+.5at^2$$

To figure out the range we have two equations with respect to time. At it's max range the displacement of the vertical component will be 0. So the first step is to find where dy = 0. It should yield two answers since dy = 0 at the beginning (t=0) and at the end of the path of the projectile since it will take a path similar to a parabola.

$$0 = t(v_{o}\sin{A}+.5at)$$
$$t = \frac{-2v_{o}\sin{A}}{a}$$

To figure out the range (distance in the horizontal direction) we need to know the time (which we just solved).

$$d_x = v_{o}\cos{A}(\frac{-2v_{o}\sin{A}}{a})$$
$$d_x = \frac{(-v_{o}^2)(2\sin{A}\cos{A})}{a}$$

That's a double-angle identity:
$$\sin{2A} = 2\sin{A}\cos{A}$$

So...

$$d_x = \frac{-v_{o}^2\sin{2A}}{a}$$?

------------
That was actually sorta fun XD

Last edited: Jun 17, 2007
7. Jun 17, 2007

### turdferguson

Yes, thats the "range equation", assuming no air resistance. Notice that complementary angles result in the same range.

As for the original question, you dont need calculus for a parabola. Find the axis of symmetry and plug in to find the y coordinate of the vertex.

8. Jun 17, 2007

### chaoseverlasting

g is 9.8m/s^2 (approx)

9. Jun 17, 2007

### threetheoreom

yes it was a mistake

yes thats right it 32ft/s^2 ..or 9.8m/s^2

10. Jun 17, 2007

### threetheoreom

Yes thats gives a elegant solution ..but it is really simpler than that ..jusy realizing the zero and solving fo it using the the g.c as a constant. like the first first responder hinted/ showed.

11. Jun 17, 2007

### HallsofIvy

Staff Emeritus
All I can say is that you should rethink your "reasoning". Why should the height be 0 at the greatest height? Are you saying it is impossible to have positive height? Here's how you should be reasoning- as long as the velocity is positive, the object is still going up- it is not yet at its greatest height. If the velocity is negative, it is already comming down- it is past its greatest height. The greatest height must occure when the velocity is 0! Since you are given that the height is h = vt-1/2gt^2 do you have a formula for velocity at time t? (if the initial velocity is v and the acceleration is -g, what is the velocity after time t?) Set that equal to 0 and solve for the time at which the object is at its greatest height. Then put that into the
formula for height.

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