h=-16t^2+22t+160

Question

a ball is thrown straight up from a rooftop 160 feet high. the formula below describes the ball's height above the ground, h, in feet, t seconds after it was thrown. the ball misses the rooftop on it's way down. the graph of the formula is shown determine when the ball's height will be 140 feet and identify the solution as a point on the graph h=-16t^2+22t+160

Answer 1

140 = -16t² + 22t + 160

16t² - 22t -20 = 0

16t² - 22t -20 = 0

Dividing both sides by 2 yields:

8t² - 11t - 10 = 0

 

Using Quadratic Eqn:

11 ± √[11² - 4(8)(-10)]

----------------------------

2(8)

 

11 ± 21

----------

    16

 

32/16   or  -10/16

 

t=2      We throw out the negative solution because it makes no sense to have a negative time in this problem.  

The ball will be at 140 ft 2 seconds after it is thrown.

 

We (you) can keep the negative solution in mind when drawing the graph :)

 

Answer 2

h(t) = -16t² + 22t + 160

 

We need to find when the ball will be 140 ft above the ground so solve for t when h = 140

 

140 = -16t² + 22t + 160

 

-16t² +22t+ 20 = 0

We can factor a 2 out of everything giving...

2(-8t²+ 11t + 10) = 0

-8t²+ 11t + 10 = 0

Multiply thru by -1 so the squared term is positive (I find it easier that way)

8t² - 11t -10 = 0

 

To factor by grouping multiply the coefficient of the square term by the constant..

(8)(-10) = -80

Look for factors of -80 that add to the coefficient of the t term which is -11

The factors of -80 that add to -11 are -16(5)

Split the -11t into -16t + 5t and substitute back into the equation

 

8t²-16t+5t-10 = 0

 

Group into two sets...   (8t²-16t) + (5t-10) = 0

Factor both sets of parentheses

8t(t-2) + 5(t-2) = 0

(8t+5)(t-2) = 0

 

Either 8t+5 = 0         or     t-2 = 0

          8t = -5                   t=2

           t = -5/8

 

We can disregard the negative answer because we know we are

looking for the amount of time it took for the ball to go up to

the apex of its trajectory then fall back to a height of 140 ft.

 

t = 2

 

The ball will reach 140 ft 2 seconds after it is thrown.

 

Check answer:  -16(2²) + 22(2) + 160 = -64 + 44 + 160

                         = 204-64 = 140 ft

This is correct

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To plot this as a point on the graph we draw a normal 4 region Cartesian coordinate plane

Label the horizontal axis as "t" for time.  Label the vertical axis as "h" for height.

Plot the point (t,h) as (2,140) on the graph.  I would suggest labeling you h axis in

groups of 10ft so that you go up 10, 20, 30, etc... to avoid having to draw an axis all

the way from 1 to 140.