SOLUTION: Hi. My teacher gave me this problem and i dont understand it: The general formula for the height, h, after t seconds of a projectile launched upward from the ground level with i

Question 167705: Hi. My teacher gave me this problem and i dont understand it:
The general formula for the height, h, after t seconds of a projectile launched upward from the ground level with initial speed, v is h=vt-1/2gt^2 where g is the gravatational constant (9.8 m/s^2 or 32 ft/s^2). Show that the greatest height of the projectiles is v^2/2g.
thanks.
Found 2 solutions by jim_thompson5910, Earlsdon:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
It might be hard to see, but the equation can be rearranged to and fits the form where , , and

So to find the greatest height, we need to find the vertex.

To do that, we use this formula

Plug in ,

Multiply and reduce

Reduce

So the max height will occur when the time is

---------------------------------

Go back to the original equation

Plug in

Square to get

Cancel like terms. (one pair of "g" terms cancel)

Multiply

Multiply the first fraction by

Combine the fractions

Subtract

So at the time the max height will be
Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
Prove that the maximum height of the projectile launched upward is: .
Starting with the function for the height (as a function of time, t) of a projectile launched upward from an initial height of with an initial velocity of :

In the given problem:
Projectile is launched from the ground, so you have:

When graphed, this equation describes a parabola that opens downward (negative coefficient of so the vertex is a maximum.
The abscissa (corresponds to the x-coordinate) of the vertex is given by:
where the a and b come from the general form of a quadratic equation:, but, in this problem, the independent variable is t rather than x, and and , so you have:
This is the time at which the projectile will be at its maximum height. Let's call this . Simplifying this, you get:
Now substitute this value of t into the original equation to find the maximum height of the projectile:
Simplifying:

QED
The maximum height attained by the projectile in this problem is:

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