ODE4.4

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A ball is thrown straight up from the ground. How high will it go?

The height of the ball at time $t\,$ is $x(t)\,$ , and the force acting on the ball is $-mg\,$ .

Newton's second law gives the relation

$m \frac{d^2x}{dt^2} + mg = 0\,$

Let $m=1\,$ for convenience.

$\frac{d^2x}{dt^2} + g = 0\,$

Integrate the last equation.

$\frac{dx(t)}{dt} + gt + C = 0\,$

To get $C\,$ , start with an initial velocity $v_0\,$ and set $t=0\,$

$v_0 + C = 0\,$

So now,

$\frac{dx}{dt} + gt - v_0 = 0\,$

Integrate again.

$x(t) + \frac{1}{2}gt^2 - v_0t + D = 0\,$

Since we have set $x(0) = 0\,$ , $D=0\,$

The equation of position is

$x(t) = -\frac{1}{2}gt^2 + v_0t\,$

To find the maximum, set the first derivative equal to 0.

$-gt+v_0=0\,$

$t_{max} = \frac{v_0}{g}\,$

Check that the second derivative at this point is negative.

$x''\left(\frac{v_0}{g}\right) = -g\,$

So the answer is this point $t_{max}\,$ evaluated in the position function.

$x_{max} = x\left(\frac{v_0}{g}\right) = \frac{v_0^2}{2g}\,$

Ordinary Differential Equations

Calculus

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