Question

A golf ball is hit from the ground with an initial velocity of 208 ft./s. assume the starting height of the ball is 0 feet. how long will it take the golfball to hit the ground?
13 sec
21 sec
42 sec
6 sec

Answer 1

Check the picture below, it hits the ground at y = 0

thus

$\bf \qquad \textit{initial velocity}\\\\ \begin{array}{llll} \qquad \textit{in feet}\\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\textit{initial velocity of the object}\\ h_o=\textit{initial height of the object}\\ h=\textit{height of the object at "t" seconds}\\ ----------\\ v_o=208\\ h_o=0 \end{cases}$

$\bf h(t)=-16t^2+208t+0\implies h(t)=-16t^2+208t\impliedby \begin{array}{llll} \textit{now let's}\\ make\\ h(t)=0 \end{array} \\\\\\ 0=-16t^2+208t\implies 0=t(208-16t) \\\\\\ \begin{cases} 0=t\\ ----------\\ 0=208-16t\\ \qquad 16t=208\\ \qquad t=\frac{208}{16} \end{cases}$

and I'm sure you know how much that is

it hits it at two points, as you notice in the picture, at 0 seconds, when it's first hit and goes up, and when it falls down.