Question

an object is launched 19.6 m/s from a 58.8 m tall platform. when does the object strike the ground? the equation for the objects height s at time t seconds after launch is s(t) = -4.9t² + 19.6t + 58.8.

Explanation:

Step1: Identify the kinematic - equation

The general equation for vertical displacement is $s = s_0+v_0t-\frac{1}{2}gt^2$, where $s = 0$ (ground - level), $s_0=58.8$ m (initial height), $v_0 = 19.6$ m/s (initial velocity), and $g = 9.8$ m/s² (acceleration due to gravity). So, $0=58.8 + 19.6t-4.9t^2$.

Step2: Rearrange the equation

Divide the entire equation by $4.9$ to simplify: $0 = 12 + 4t - t^2$. Rearrange it to the standard quadratic - form $t^2-4t - 12=0$.

Step3: Solve the quadratic equation

For a quadratic equation $ax^2+bx + c = 0$ (here $a = 1$, $b=-4$, $c=-12$), the quadratic formula is $t=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$. Substitute the values: $t=\frac{4\pm\sqrt{(-4)^2-4\times1\times(-12)}}{2\times1}=\frac{4\pm\sqrt{16 + 48}}{2}=\frac{4\pm\sqrt{64}}{2}=\frac{4\pm8}{2}$.

Step4: Find the valid solution for time

We get two solutions for $t$: $t_1=\frac{4 + 8}{2}=6$ and $t_2=\frac{4-8}{2}=-2$. Since time cannot be negative in this context, we reject $t_2$.

Answer:

$t = 6$ s