Answer
62.296 m/s
Work Step by Step
First use the acceleration function to find the velocity and position functions.
$a(t) = -9.8$
$\int a(t) = v(t) = -9.8t + v_0$
$\int v(t) = s(t) = -4.9t^2 + v_0t + s_0$
Since the object is thrown from a height of 2 meters, $s_0 = 2$. Therefore:
$s(t) = -4.9t^2 + v_0t + 2$
The maximum height will occur when $v(t) = 0$, as velocity is the derivative of position. Therefore, find the time at which $v(t) = 0$, in terms of $v_0$.
$0 = -9.8t + v_0$
$9.8t = v_0$
$t = \frac{v_0}{9.8}$
Since the maximum height needs to be 200, $s(t) = 200$ at time $t = \frac{v_0}{9.8}$.
$200 = -4.9(\frac{v_0}{9.8})^2 + v_0(\frac{v_0}{9.8}) + 2$
Solve for $v_0$.
$198 = -4.9(\frac{v_0^2}{9.8^2}) + \frac{v_0^2}{9.8}$
$198 = -4.9(\frac{v_0^2}{9.8^2}) + 9.8(\frac{v_0^2}{9.8^2})$
$198 = 4.9(\frac{v_0^2}{9.8^2})$
$40.408 = \frac{v_0^2}{9.8^2}$
$3880.8 = v_0^2$
$v_0 = 62.296$ m/s
