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

Step-by-step explanation:

Equation is given by p(t) = 3gt² + v₀t + P₀

Given that

                g = -32 ft/s²

                v₀ = 24 ft/s

                P₀ = 5 ft

Substituting

              p(t) = 3 x -32 x t² + 24 x t + 5

              p(t) = -96 t² + 24 t + 5

We need to find maximum of this equation, at maximum we have derivative is zero.

              p'(t) = -96 x 2t + 24 = 0

             192 t = 24

                    t = 0.125 s

Substituting in p(t) equation

              p(0.125) = -96 x 0.125² + 24 x 0.125 + 5

              p(0.125) = 6.5 ft

Maximum height reached is 6.5 ft

he earned 78.75 in interest.

step-by-step explanation: