A baseball is thrown in a parabolic arc. It's position above the ground at a given point in timecan be represented by the quadratic function p(t) = 3gt2 + vot + Po, where t > 0,9 is -32ft/sec/sec, vo is initial velocity, and Po is its initial position above the ground. If the ball wasthrown straight up at 24 ft/sec when it was 5 ft above the ground, how high did it go? (Type

If the ball was thrown straight up at 24 ft/sec when it was 5 ft above the ground, the ball reached a maximum height of 7.25 m[tex]\texttt{ }[/tex]Further explanationDiscriminant of quadratic equation ( ax² + bx + c = 0 ) could be calculated by using :D = b² - 4 a cFrom the value of Discriminant , we know how many solutions the equation has by condition :D < 0 → No Real RootsD = 0 → One Real RootD > 0 → Two Real Roots[tex]\texttt{ }[/tex]An axis of symmetry of quadratic equation y = ax² + bx + c is :[tex]\large {\boxed {x = \frac{-b}{2a} } }[/tex]Let us now tackle the problem![tex]\texttt{ }[/tex]Given:[tex]p(t) = 3gt^2 + v_ot + p_o[/tex][tex]p(t) = 3(-32)t^2 + 24t + 5[/tex][tex]p(t) = -64t^2 + 24t + 5[/tex]Asked:[tex]p_{max} = ?[/tex]Solution:At the maximum height , velocity is 0 m/s:[tex]v = \frac{dp(t)}{dt}[/tex][tex]v = \frac{d}{dt} ( -64t^2 + 24t + 5 )[/tex][tex]v = (-64)(2)t^{2-1} + 24[/tex] [tex]v = -128t + 24[/tex][tex]0 = -128t + 24[/tex][tex]128t = 24[/tex][tex]t = 24 \div 128[/tex][tex]t = 3 \div 16[/tex][tex]t = 0.1875 \texttt{ s}[/tex][tex]\texttt{ }[/tex][tex]p(t) = -64t^2 + 24t + 5[/tex][tex]p(0.1875) = -64(0.1875)^2 + 24(0.1875) + 5[/tex][tex]p(0.1875) = 7.25 \texttt{ m}[/tex][tex]\texttt{ }[/tex]Learn moreSolving Quadratic Equations by Factoring : the Discriminant : of Quadratic Equations : [tex]\texttt{ }[/tex]Answer detailsGrade: High SchoolSubject: MathematicsChapter: Quadratic Equations[tex]\texttt{ }[/tex]Keywords: Quadratic , Equation , Discriminant , Real , Number#LearnWithBrainly