Question

On the moon, the height of a golf ball hit with an initial velocity of 96 ft/sec is given by the equation h= -2.7t^2+96t, where h represents the height of the ball in feet after t sec. On Earth, the same shot would be modeled by the equation h= -16t^2+96t. How much longer will the flight time be on the moon than on the Earth?

Answer 1

So hmmm notice the picture below

"x" being how many seconds the object was going

and when y = 0, the object hits the ground, either on the moon or on earth

so, if we set y = 0 on each equation, we'll know when that happen, how long of a "x-value" or seconds it took

$\bf \begin{array}{llll} \textit{on the moon}\\\\ h=-2.7t^2+96t\implies 0=t(96-2.7t) \end{array} \quad \begin{cases} 0=t\\ -------\\ 0=96-2.7t\\ 2.7t=96\\ t=\frac{96}{2.7}\\ t\approx 35.\overline{55} \end{cases}\\\\ -------------------------------\\\\ \begin{array}{llll} \textit{on earth}\\\\ h=-16t^2+96t\implies 0=t(96-16t) \end{array} \quad \begin{cases} 0=t\\ ------\\ 0=96-16t\\ 16t=96\\ t=\frac{96}{16}\\ t=6 \end{cases}$