Question 140056
h=-5t^2 +15t +45


Yes, the height of the building is the height at t=0, which is 45 ft.


Maximum height of the parabola {{{y=ax^2 +bx +c}}} occurs at {{{x=-b/(2a)}}}

Maximum height at ={{{t=-15/(2*-5)=15/10=3/2}}} seconds.
Maximum height = {{{-5*(3/2)^2+15*(3/2)+45}}} feet
Maximum height = {{{-5(9/4)+45/2+45=-45/4+45/2+45}}}
Maximum height = {{{-45/4+90/4+45=45/4+45=11.25+45 =56.25}}} feet


The ball reaches 55 feet when 
{{{55=-5t^2 +15t +45}}}
{{{0=-5t^2+15t-10}}}
{{{0=-5(t^2-3t+2) }}}
{{{0=-5(t-1)(t-2) }}}

t=1 and t=2 seconds (at t=1 sec going up, at t=2 seconds coming down!)


The ball hits the ground when h=0.


h=-5t^2 +15t +45
0=-5t^2 +15t +45
0=-5(t^2-3t-9)
Solve this by quadratic formula, since it does not factor.  The answer will contain a radical.


R^2