Algebra II

Question

The function h = -16t2 + 16t + 410 gives the height h, in feet, of a coin tossed upward from a balcony 410 feet high with an initial velocity of 16 ft/sec. During what interval of time will the coin be at a height of at least 90 feet?

Question 17 options:

1) [4, 6]

2) [6, 5]

3) [0, 7]

4) [0, 5]

Answer 1

Hi Carrie.  The easiest way to do this is by the process of elimination (much like one would on the SATs).  We can immediately eliminate 2) [6,5] as the begin time of the interval is greater than the end time.  Next, if we plug the endpoints of each interval into the height equation we can get a sense of the height of the coin at each.  Let's try 1) [4,6]:

 

h(4) = -16(4)² + 16(4) + 410 = 218 

h(6) = -70 (I will eliminate the details from here on)

 

We can see that although the coin started out at a height > 90 it ended up at a height of -70 (actually 70 ft below ground).  It does not matter for how much of the interval it was below 90 ft.  The way the question is worded demands that the coin be at a height >= 90 during the entire interval.  

 

We can use the same method for 3) [0,7].  Here are the heights of the interval endpoints:

 

h(0) = 410

h(7) = -262

 

Using the same reasoning as for 2) we can say that the coin during this time was NOT at or above 90 feet during the entire interval.

 

If this were the SATs I would advise you to stop and immediately bubble in the 4) answer, as all the other options cannot be true and time during that test is of the essence.  However, as a check I will do the same analysis for option 4:

 

h(0) = 410

h(5) = 90

 

We may be tempted to immediately say that this is a correct answer since the endpoints are both >= 90.  However, in general, it is very possible for the height to dip below 90 at some point within this range and then make it back up to 90 at t=5.  A little knowledge of quadratic equations (one where the highest exponent is 2) tells us that this cannot happen for this equation during this interval so the answer is in fact 4.

 

 

Answer 2

h  = -16 t² + 16 t + 410 

h = 90 ft, t = ?

 

90 = -16 t² + 16 t + 410 

-16 t² + 16 t + 410 = 90

-16 t² + 16 t + 410 - 90 = 0

-16 t² + 16 t + 320 = 0

 

(-16/-16) t² + (16/-16) t +( 410/-16) = 0/-16

t² - t - 20 = 0

 

t = (b² (+) square root (b² - 4ac))/(2a)

 

from our equation, 

a = 1, b = -1, c = -20

 

using quadratic formula, 

 

t = (1 (+) square root ( 1 - 4(1)(-20))/(2)

t = (1 (+) square root ( 1 +80))/2

t = (1 (+) square root ( 81))/2

t = (1 (+) 9)/2

t = (10/2 or -8/2)

t = (5 or -4)

since time cant be negative t = 5 sec at 90 ft 

 

When the coin is at 410 ft, the time is 0 because that is its initial point 

 

Therefore, the interval is [0, 5]