High school maths

Question

An arrow is shot upward with a velocity of 64 feet per second. Using the formula         h(t)=64t-16t^2 how long after the arrow is released does it hit the ground?

Answer 1

h(t) = 64t - 16t²        formula

0 = 64t - 16t²            substitute height 

0 = 16t(4 - t)            factor our GCF, 16t

16t = 0                     zero product rule

t = 0

or

0 = 4 - t

t = 4

 

When does the arrow return to the ground?

 

 

Answer 2

First, read and re-read the problem until you understand it and can put it into your own words.  How's this:

    h(t) = 64t-16t²        What is t when h=0  (the ground)?  

 

    0 = 64t -16 t²

    0 = 16t(4-t)

 

The multiplicative property of zero says that one or all of the factors must equal 0, so either:

     t = 0           (that's the start time)

  or

     t = 4 seconds

Answer 3

h(t) = 64t - 16t² = t(64 - 16t)

 

The arrow hits the ground when h = 0

 

h(t) = t(64 - 16t) = 0 ⇒ t = 0 or t = 64/16 = 4s

 

The t = 0 solution is the initial moment when the arrow is shot up from the ground at h = 0.  The next time the arrow is at the ground is t = 4s.