Question 724451
If you use the formula correctly then you can't get the wrong answer.<br>
{{{0 = -4.9t^2 + 80t}}}
a = -4.9
b = 80
c = 0
{{{t = (-(80)+-sqrt((80)^2-4(-4.9)(0)))/2(-4.9)}}}
{{{t = (-(80)+-sqrt(6400-4(-4.9)(0)))/2(-4.9)}}}
{{{t = (-(80)+-sqrt(6400-0))/2(-4.9)}}}
{{{t = (-(80)+-sqrt(6400))/2(-4.9)}}}
{{{t = (-(80)+-80)/2(-4.9)}}}
{{{t = (-80 +- 80)/(-9.8)}}}
which is short for:
{{{t = (-80 + 80)/(-9.8)}}} or {{{t = (-80 - 80)/(-9.8)}}}
{{{t = 0/(-9.8)}}} or {{{t = (-160)/(-9.8)}}}
{{{t = 0}}} or {{{t = 800/49}}} (or approximately 16.326530612244897959183673469388)
The answer of zero reflects the fact that the object started on the ground. The other answer is the one you are looking for.<br>
Of course this problem did not need the quadratic formula. The right side of
{{{0 = -4.9t^2 + 80t}}}
factors!
{{{0 = t(-4.9t + 80)}}}
From the Zero Product Property:
t = 0 or -4.9t + 80 = 0
Solving the second one:
t = 0 or t = 800/49<br>
It's much easier to solve these if you can factor them. The formula is complicated and mistakes can easily be made. (You must have made a mistake yourself if you didn't get the right answer.)