SOLUTION: Solve by completing the square A ball is thrown upward at an initial velocity of 26 ft/sec^2 from a height of 3 ft. Find the approximate length of time(to the nearest tenth of a

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: Solve by completing the square A ball is thrown upward at an initial velocity of 26 ft/sec^2 from a height of 3 ft. Find the approximate length of time(to the nearest tenth of a      Log On


   

Question 1028882: Solve by completing the square
A ball is thrown upward at an initial velocity of 26 ft/sec^2 from a height of 3 ft. Find the approximate length of time(to the nearest tenth of a second) it will take the ball to return to a height of 5 ft.
USE THIS FORMULA:
h(t)= -16t^2 + vt + h
Where V(measured in ft/sec^2) and H(initial height)
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!

There are 2 times when the ball's height is 5 ft.
On the way up and on the way down.
So if we plug in 5 for h, we should get 2 values for t
5 = 3 +26t -16t^2
16t^2-26t+2=0
Find the roots of the equation by quadratic formula

a= 16 b= -26 c= 2

b^2-4ac= 676 - 128
b^2-4ac= 548 sqrt%28%09548%09%29= 23.41
x=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29
x1=%28-b%2Bsqrt%28b%5E2-4ac%29%29%2F%282a%29
x1=( 26 + 23.41 )/ 32
x1= 1.54
x2=( 26 -23.41 ) / 32
x2= 0.08

t= 0.08 OR 1.54