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jek_recluse [69]
4 years ago
10

A person standing close to the edge on top of a 24-foot building throws a ball vertically upward. The quadratic function h(t)=−1

6t^2+92t+24 models the ball's height about the ground, h(t), in feet, t seconds after it was thrown.
a) What is the maximum height of the ball?

__feet

b) How many seconds does it take until the ball hits the ground?

____ seconds
Mathematics
1 answer:
Nutka1998 [239]4 years ago
8 0

Answer:

a) maximum height is 156.25 ft

b) 6 seconds

Step-by-step explanation:

a) The maximum height can be found by finding the y-coordinate of the vertex. The vertex is the highest point in a parabola opened downward.

I start with by finding the t-coordinate of the vertex.

The t-coordinate of the vertex is \frac{-b}{2a} where the expression -16t^2+92t+24 will need to be compared to at^2+bt+c.

We see that a=-16,b=92,c=24.

So the t-coordinate of the vertex is \frac{-b}{2a}=\frac{-92}{2(-16)}=\frac{-92}{-32}=\frac{23}{8}.

We can find the h, the height, that corresponds to this t by using the equation h=-16t^2+92t+24 where t=\frac{23}{8}.

So inserting 23/8 for t.

-16(23/8)^2+92(23/8)+24

Plugging into calculator gives you 625/4.

So the maximum height is 625/4 or 156.25.

b) Let's find how long it takes the ball to hit the ground. If the ball is on the ground, then the distance between the ball and the ground is 0.  So we are looking to solve h(t)=0 for t.

So this is equation we are solving for t:

-16t^2+92t+24=0

These numbers are big but since all the terms have a common factor we can make them slightly smaller.

That is, we are going to divide both sides by -4 and see

4t^2-23t-6=0

It looks like it could be possible to factor.

a=4

b=-23

c=-6

We need to find two numbers that multiply to be ac and add up to be b.

a*c=4(-6)=-24

b=-23

So -24=-24*1 and -23=-24+1.

We are going to replace -23t with -24t+1t giving up something that should be factorable by grouping.

4t^2-23t-6=0

4t^2-24t+1t-6=0

Group the first 2 together and group the last two together:

(4t^2-24t)+(1t-6)=0

Factor what you can from each pair:

4t(t-6)+1(t-6)=0

Now each term has a common factor of (t-6) so factor that out giving you:

(t-6)(4t+1)=0

Set both factors equal to 0 giving you

t-6=0              or               4t+1=0

  t=6               or               4t  =-1

  t=6               or                 t=-1/4

So it hit the ground in 6 seconds.

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Consider the lengths of consecutive 1-2 blocks.

block 1 - 1, 2 - length 2

block 2 - 1, 2, 2 - length 3

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and so on.


Recall the formula for the sum of consecutive positive integers,

\displaystyle \sum_{i=1}^j i = 1 + 2 + 3 + \cdots + j = \frac{j(j+1)}2 \implies \sum_{i=2}^j = \frac{j(j+1) - 2}2

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which means that the 1234th term in the sequence occurs somewhere about 1/5 of the way through the 49th 1-2 block.

In the first 48 blocks, the sequence contains 48 copies of 1 and 1 + 2 + 3 + ... + 47 copies of 2, hence they make up a total of

\displaystyle \sum_{i=1}^48 1 + \sum_{i=1}^{48} i = 48+\frac{48(48+1)}2 = 1224

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\displaystyle \sum_{i=1}^{48} 1 + \sum_{i=1}^{48} 2i = 48 + 48(48+1) = 48\times50 = 2400

This leaves us with the contribution of the first 10 terms in the 49th block, which consist of one 1 and nine 2s with a sum of 1+9\times2=19.

So, the sum of the first 1234 terms in the sequence is 2419.

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