A ball is launched upward from a height 40 feet above ground level. The ball’s height at t seconds is given by -16t^2=128=40 .

Answer:The ball will be at 100 feet at time x=0.5 sec and at time x=7.5 secStep-by-step explanation:The correct question isA ball is launched upward from a height 40 feet above ground level. the ball's height at t seconds is given by h(t)=-16t^2+128t+40 At what time(s) will the ball be at 100 feet?we have[tex]h(t)=-16t^{2}+128t+40[/tex]soFor h(t)=100 ftsubstitute in the equation and solve for x[tex]-16t^{2}+128t+40=100[/tex][tex]-16t^{2}+128t-60=0[/tex]The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to [tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex] in this problem we have [tex]-16t^{2}+128t-60=0[/tex]so [tex]a=-16\\b=128\\c=-60[/tex] substitute in the formula [tex]x=\frac{-128(+/-)\sqrt{128^{2}-4(-16)(-60)}} {2(-16)}[/tex] [tex]x=\frac{-128(+/-)\sqrt{12,544}} {-32}[/tex] [tex]x=\frac{-128(+/-)112} {-32}[/tex] [tex]x=\frac{-128(+)112} {-32}=0.5[/tex] [tex]x=\frac{-128(-)112} {-32}=7.5[/tex] thereforeThe ball will be at 100 feet at time x=0.5 sec and at time x=7.5 sec