Question

Suppose you throw a baseball into the air with an initial upward velocity of 29 ft/s and an initial height of 6ft. The formula h(t) = -16t ^2 + 29t + 6 (^ means to the power of. So exponents), gives the ball’s height, h, in feet at time, t, in seconds. Find the number of seconds that pass before th ball hits the ground!

Answer 1

Check the picture below.

$\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in feet} \\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{29}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{6}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\\\ h(t)=-16t^2+29+6\implies \stackrel{h(t)}{0}=-16t^2+29+6 \\\\\\ 16t^2-29-6=0\implies (16t+3)(t-2)=0\implies t= \begin{cases} -\frac{3}{16}\\\\ \boxed{2} \end{cases}$

it cannot be a negative value, since it's seconds after the ball went up, so is not -3/16.