Question 112161
The height of an object propelled upwards from an initial height of {{{h[0]}}} with an initial velocity of {{{v[0]}}} is given by the function (height as a function of time):
{{{h(t) = -16t^2+v[0]t+h[0]}}}
In your problem:
{{{v[0] = 20}}}ft/sec. and
{{{h[0] = 120}}}ft.
You want to find out at what time does the height (h) = 0, so, making the appropriate substitutions into the equation, we get:
{{{0 = -16t^2+20t+120}}}
Using the quadratic formula to solve:
{{{t = (-20+-sqrt(20^2-4(-16)(120)))/2(-16)}}} Simplifying this, we get:
{{{t = (-20+-sqrt(400+7680))/-32}}}
{{{t = (-20+-sqrt(8080))/-32}}}
{{{t = (-20+89.9)/-32}}} or {{{t = (-20-89.9)/-32}}}
{{{t = -2.184}}} or {{{t = 3.434}}}
Only the positive value is meaningful here, so the ball reaches the ground in 3.434 seconds.