Question

An object is launched upward

from 62.5 meters above ground level with an
initial velocity of 12 meters per second. The
gravitational pull of the earth is about 4.9
meters per second squared. How long will the
object take to hit the ground?​

Answer 1

Check the picture below. The picture is using feet, but is pretty much the same trajectory for meters.

so the object hits the ground when y = 0, thus

`\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in meters} \\\\ h(t) = -4.9t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{12}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{62.5}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at`

`\bf ~~~~~~~~~~~~\textit{using the quadratic formula} \\\\ h(t)=\stackrel{\stackrel{a}{\downarrow }}{-4.9}t^2\stackrel{\stackrel{b}{\downarrow }}{+12}t\stackrel{\stackrel{c}{\downarrow }}{+62.5} \qquad \qquad t= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a} \\\\\\ t=\cfrac{-12\pm√(12^2-4(-4.9)(62.5))}{2(-4.9)}\implies t=\cfrac{-12\pm√(144+1225)}{-9.8} \\\\\\ t=\cfrac{12\mp√(1369)}{9.8}\implies t=\cfrac{12\mp 37}{9.8}\implies t= \begin{cases} \stackrel{\approx}{-2.55}\\ 5 \end{cases}`

we can't use -2.55, since it's seconds and can't be negative, so t = 5 seconds later.