Question 193178
I think the key is that {{{h}}} is the height from the ground and the
problem wants to know when the penny reaches ground, or when
{{{h = 0}}}
{{{h = -16t^2 - vt + s}}}
{{{0 = -16t^2 - vt + s}}}
given:
{{{v = 12}}} ft/sec
{{{s = 46.75}}} ft
Another key is that signs are already figured into the
equation, so I don't have to worry about signs
--------------------
{{{ -16t^2 - 12t + 46.75 = 0}}}
Using the quadratic formula:
{{{t = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{a = -16}}}
{{{b = -12}}}
{{{c = 46.75}}}
{{{t = (-(-12) +- sqrt( (-12)^2-4*(-16)*46.75 ))/(2*(-16)) }}}
{{{t = (12 +- sqrt( 144 + 2992 ))/(-32) }}}
{{{t = (12 +- 56)/(-32) }}}
If the numerator is (+), then {{{t}}} is negative, and I can't have
negative time, so the numerator must be negative
{{{t = (-44)/ (-32)}}}
{{{t = 1.375}}} sec
The penny will reach the ground in 1.375 seconds
check answer:
{{{ -16t^2 - 12t + 46.75 = 0}}}
{{{(-16)*(1.375)^2 - 12*1.375 + 46.75 = 0}}}
I don't have a calculator, but I think this checks out