If the ball was thrown straight up at 24ft/sec when it was 5 ft above the ground, how high did it go?

Question

A baseball 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)=1/2gt^2 +v0t +p0, where t is greater than or equal to 0, g is -32 ft/sec/sec, v0 is initial velocity, and p0 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?

Answer 1

p(t) = -16t^2 +24t + 5

take the derivative, set = 0 and solve for t

p'(t) = -32t +24 = 0

t = 24/32 = 3/4 seconds to reach max height

p(3/4) = -16(3/4)^2 + 24(3/4) + 5 = 14 feet = max height