Question 167705
Prove that the maximum height of the projectile launched upward is: {{{v[0]^2/2g}}}.
Starting with the function for the height (as a function of time, t) of a projectile launched upward from an initial height of {{{h[0]}}} with an initial velocity of {{{v[0]}}}:
{{{h(t) = -(1/2)gt^2+v[0]t+h[0]}}}
In the given problem:
{{{h[0] = 0}}} Projectile is launched from the ground, so you have:
{{{h(t) = -(1/2)gt^2+v[0]t}}}
When graphed, this equation describes a parabola that opens downward (negative coefficient of {{{t^2}}} so the vertex is a maximum.
The abscissa (corresponds to the x-coordinate) of the vertex is given by:
{{{x = -b/2a}}} where the a and b come from the general form of a quadratic equation:{{{f(x) = ax^2+bx+c}}}, but, in this problem, the independent variable is t rather than x, and {{{a = -(1/2)g}}} and {{{b = v[0]}}}, so you have:
{{{t = -v[0]/2(-(1/2)g)}}} This is the time at which the projectile will be at its maximum height. Let's call this {{{t[m]}}}. Simplifying this, you get:
{{{t = v[0]/g}}} Now substitute this value of t into the original equation {{{h(t) = -(1/2)gt^2+v[0]t}}} to find the maximum height of the projectile:
{{{h(t[m]) = -(1/2)g(v[0]/g)^2+v[0](v[0]/g)}}} Simplifying:
{{{h(t[m]) = -(1/2)g(v[0]^2/g^2)+v[0]^2/g}}}
{{{h(t[m]) = -v[0]^2/2g + v[0]^2/g}}}
{{{h(t[m]) = v[0]^2/g}}} QED 
The maximum height {{{h[m]}}} attained by the projectile in this problem is:
{{{highlight(h[m] = v[0]^2/2g)}}}