Solve the following application problem (show your equation and then your solution). Round answers to nearest?

Question

If a ball is thrown vertically upward from a height of 6 ft, with an initial velocity of 64 ft/s, its height h after t seconds is given by h = -16t^2 + 64t + 6. How long does it take the ball to reach a height of 38 ft on the way up?

Here's what I have so far and I'm stumped.

-16t^2+64t+6=38 subtract 38 from both sides
-16t^2+64t-32=0 factor out the 16
16(t^2-4t+2)=0 now what?

Answer 1

SOlve the quadratic: t^2 - 4t + 2 = 0

It isnt factorable so you need to use the quadratic formula.

Discriminant = b^2 - 4ac = 16 - 4(1)(2) = 8

t =(4 + sqrt 8)/2 or t = (4 - sqrt8)/2

t= 3.41 or t = 0.59

Since you are looking for on the way up it will be the 0.59 s.

Answer 2

Divide by 16 on both sides..

t^2 - 4t + 2 = 0

You can't factor it, so use the quadratic formula.
You get 2 +/- rt.2

Use your calculator to find an approximate solution:

.5858 and 3.414

On the way up would be the first one.. about .5858 seconds =)

Answer 3

h=-16t^2+112t h=one hundred and eighty -16t^2+112t=one hundred and eighty 16t^2-112t+one hundred and eighty=0 Divide it by way of four: 4t^2-28t+40 5=0 (2t-5)(2t-9)=0 t=2.5 or 4.5 The arrow is going as a lot as one hundred and eighty feet. in 2.5 seconds and four.5 seconds on it truly is way down.