Question 92924
When the object will reach ground, then h = 0.
So {{{-16t^2 + 60t + 9 = 0}}}
{{{16t^2 - 60t - 9 = 0}}}


Comparing with the standard form {{{ax^2 + bx + c = 0}}}, here a = 16, b = -60, c = -9.
From Sreedharacharya's formula
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{t = (-(-60) +- sqrt( (-60)^2-4*16*(-9) ))/(2*16) }}} [substituting the values of a, ,b & c]
{{{t = (60 +- sqrt( 3600+576 ))/32 }}}
{{{t = (60 +- 64.622)/32 }}}
{{{t = 3.894}}} or {{{t = -0.145}}}


But {{{t}}} is time and hence cannot be negative.
So the object will reach ground after {{{t = 3.894}}} seconds.