Janell is standing on a set of bleachers and throws a ball into the air at an initial velocity of35 ft/s. The height of the ball, h, at 1 seconds is modeled by the equation h = -162 +35t + 6How many seconds will it take the ball to reach the ground

Answer:The time taken to reach ground is 3.45 secStep-by-step explanation:Given as :Distance of ball = h = -162 + 35t + 6  at t = 1 secThe speed of ball thrown  = 35 ft per secLet the Time taken to reach ground = T secSo Time = [tex]\frac{Distance}{Speed}[/tex]Or, Time = [tex]\frac{-162+35t+6}{35}[/tex]Now distance h at t= 1 sec , h = -162 + 35t +6 or,   h = - 162 + 35(1) + 6So,  h = - 121  feetOr   [tex]\begin{vmatrix}h\end{vmatrix}[/tex] = 121 feet Or, Time = [tex]\frac{ 121}{35}[/tex] Or,  Time = 3.45 secHence The time taken to reach ground is 3.45 sec  Answer