Question 442973
Great job so far.

It will make sense when you think of it in these terms.

What height is the ground? Well, when we say we're 64 feet above the ground, and we let our s be 64, then we are saying we are 64 above 0. This implies that the ground is height 0. You may say, duh, but now we have the key to the problem.

Solve for t when h =0.

Then, {{{ 0 = 64 - 16t^2}}}

Notice that 64 and 16t^2 are perfect squares.

Recall the difference of perfect squares {{{(a^2-b^2) = (a-b)(a+b)}}}

Then, {{{64-16t^2 = (8-4t)(8+4t)}}}

Solve 8-4t = 0:

{{{8 =4t}}}
{{{t = 2}}}

Solve 8 +4t = 0
{{{8 = -4t}}}
{{{t = -2}}}

In this context, a negative time does not make sense, but a positive time does.
So the only time that makes LOGICAL SENSE is t = 2.

So {{{t=2}}}.

Hope this helped!