Question 87236
Given {{{h(t) = 73.5-4.9t^2}}} where 73.5 meters is the initial height of the rock, how long before it hits the ground?
Set h(t) = 0 and solve for t.
{{{73.5-4.9t^2 = 0}}} Add {{{4.9t^2}}} to both sides.
{{{73.5 = 4.9t^2}}} Divide both sides by 4.9
{{{15 = t^2}}} Now take the square root of both sides.
{{{t = 3.873}}}seconds.
The best choice of answers is:
3.9 seconds.