# Thread: Applications of Calculus - maximum height

1. ## Applications of Calculus - maximum height

A body is projected vertically upwards from the ground in a medium that produces a resistance force per unit mass of $\displaystyle kv^2$, where v is the velocity and k is a positive constant. Given that the maximum height is $\displaystyle H = \frac{1}{2k}ln (1+\frac{k(v_0)^2}{g})$ Show that to double the maximum height, $\displaystyle v_0$ must be increased by a factor of $\displaystyle (e^{2kH} + 1)^\frac{1}{2}$
where $\displaystyle v_0$ is the initial velocity

Thanks

2. [ΜΑΤΗ]Η = \frac{1}{2k}ln(1 + \frac{k(v_o)^2}{g}[/tex]

$\displaystyle 2kH = ln(1 + \frac{k(v_o)^2}{g}$

$\displaystyle e^{2kH} = (1 + \frac{k(v_o)^2}{g}$

$\displaystyle e^{2kH} - 1 = \frac{k(v_o)^2}{g}$ .......(1)

For 2H, the equation (1) becomes

$\displaystyle e^{4kH} - 1 = \frac{k(v_o')^2}{g}$ ........(2)

Now divide (2) by (1) and simplify to find vo'

3. $\displaystyle H = \frac{1}{2k}ln (1+\frac{k(v_0)^2}{g})$

$\displaystyle 2kH = ln(1 + \frac{k(v_o)^2}{g})$

$\displaystyle e^{2kH} = (1 + \frac{k(v_o)^2}{g})$

$\displaystyle e^{2kH} - 1 = \frac{k(v_o)^2}{g}$ .......(1)

For 2H, the equation (1) becomes

$\displaystyle e^{4kH} - 1 = \frac{k(v_o')^2}{g}$ ........(2)

Now divide (2) by (1) and simplify to find vo'