A water balloon is launched from a machine upward from a height of 10 feet with an initial velocity of 46 feet per second

Question

Solve by using Quadratic In equalites

Answer 1

h(t) = (-1/2)gt² + v₀t + h₀

 

h(t) is height at a specifice time

g = acceleration of gravity = 32 ft/sec²

t = time

v₀ = initial velocity = 46

h₀ = initial height = 10

 

h(t) = -16t² + 46t + 10

 

If you are trying to find the amount of time

it will take for the water balloon to hit the

ground then h(t) = 0

 

-16t² + 46t + 10 = 0

-8t² + 23t + 5 = 0

 

from quadratic equation

 

t = [-23 ±√(23²-4(-8)(5))]/(2(-8))

 

Once you solve for t, discard the

negative answer.  The other is

your answer.