Question

An object is thrown in the air with an initial velocity of 5 m/sec from a height of 9 m. The following equation models the height of the object in meters after t seconds. How many seconds does it take for the object to hit the ground?

Answer 1

Check the picture below.

`~~~~~~\textit{initial velocity in meters} \\\\ h(t) = -4.9t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}&5\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&9\\ \qquad \textit{of the object}\\ h=\textit{object's height}\\ \qquad \textit{at`

since we don't get any neat integer values from it hmmm let's plug that in the quadratic formula to see what we get as h(t) = 0.

`~~~~~~~~~~~~\textit{quadratic formula} \\\\ \stackrel{\stackrel{a}{\downarrow }}{4.9}t^2\stackrel{\stackrel{b}{\downarrow }}{-5}t\stackrel{\stackrel{c}{\downarrow }}{-9}=0 \qquad \qquad t= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a} \\\\\\ t= \cfrac{ - (-5) \pm \sqrt { (-5)^2 -4(4.9)(-9)}}{2(4.9)}\implies t=\cfrac{5\pm√(25+176.4)}{9.8} \\\\\\ t=\cfrac{5\pm √(201.4)}{9.8}\implies t\approx \begin{cases} 1.96~~\checkmark\\ -0.94 \end{cases}`

notice, we do not use the negative value for "t", since the seconds cannot be less than 0.