Derek kicks a soccer ball off the ground and in the air, with an initial velocity of 31 feet per second. Using the formula H(t) = −16t2 + vt + s, what is the maximum height the soccer ball reaches? (5 points) 14.2 feet 14.6 feet 15.0 feet 15.3 feet

General Idea:If we have a quadratic function of the form [tex] f(x)=ax^{2} +bx+c [/tex], then the function will attain its maximum value only if a < 0 & its maximum value will be at [tex] x=-\frac{b}{2a} [/tex].Applying the concept:Comparing the function[tex] f(x)=ax^{2} +bx+c [/tex] with the given function[tex] H(t)=-16t^{2} +31t+s [/tex], we get[tex] a = -16 [/tex], [tex] b = 31 [/tex] and [tex] c=0 [/tex] because initial height of the ball s will be 0.The maximum height of the soccer ball will occur at [tex] t=\frac{-b}{2a}=\frac{-31}{2(-16)} = \frac{-31}{-32}=0.96875 seconds [/tex]The maximum height is found by substituting [tex] t=0.96875 [/tex] in the function as below:[tex] H(0.96875)=-16(0.96875)^2+31(0.96875) [/tex]Conclusion:The maximum height the soccer ball reaches is 15.0 feet approximately!