Question 113289
The height of an object, as a function of time, t, propelled upwards from an initial height {{{h[0]}}} with an initial velocity {{{v[0]}}} is given by:
{{{h(t) =-16t^2+v[0]t+h[0]}}}
In your problem:
{{{v[0] = 96}}}ft.
{{{h[0] = 0}}}ft.... or on the ground.
You want to find the values of t for which h = 0, so substitute h=0 and solve for t.
{{{0 = -16t^2+96t}}} Factor a t.
{{{0 = t(-16t+96)}}} Apply the zero product principle.
{{{t = 0}}} and
{{{16t = 96}}} Divide both sides by 16.
{{{t = 6}}}
So the height of the projectile is 0 (on the ground) at times, t = 0 secs and t = 6 secs.