Question

A juggler throws a ball into the air from a height of 5 ft with an initial vertical velocity of 16 ft/s. The function that models the height h of the ball in feet t seconds after the ball is thrown is h(t) = -16t2 + 16t + 5. How long does the juggler have to catch the ball before it hits the ground?

Answer 1

1.25 seconds to catch the ball before it hits the ground :D hope i helped

Answer 2

Answer:

t = 2.525 seconds

Step-by-step explanation:

h(t) = -16t^2 + v*t + h0  h(t) = the height above the ground after t seconds v = 40 ft/s initial velocity  h0 = 4 ft the initial height of the object (the ball)  t = the time of the motion  h(t) = -16t^2 + 40*t + 4  we need to find t such that h(t) = 3 ft  3 = -16t^2 + 40*t + 4  -16t^2 + 40*t + 4 = 3  by solving we find

t = 2.525 seconds

You can round on your own.