Another Question!

Printable View

June 19th 2008, 06:32 PM
Nikola

Another Question!

I need to know if there is a standard method of finding the answer to this. My exam is tomorrow and my teacher gave us a set of topics to study. This is titled Projectile Motion!

The question is:

The function h=-5t^2 + 20t + 2 gives the approximate height, h meters of a trown football as a function of the time, t seconds since it was thrown. The ball hit the ground before the receiver could get near it.

a) How long was the ball in the air, to the nearest tenth of a second?
b) for how many seconds was the height of the ball at least 17 m?

The answers are:
a) 4.1
b) 2

I dont understand how to get the answers though. Sorry if there are misspelling in the question i had to rewrite it from the text book. Thanks again!
June 19th 2008, 07:08 PM
Mathnasium

Remember that in $h=-5t^2 + 20t + 2$ , h stands for the height. When the ball hits the ground, the height is 0, so plug that in for h. This gives:

$0=-5t^2 + 20t + 2$ .

Now, solve that by the quadratic equation to figure out what t equals when h is zero.

Similarly, you can find out the two times the ball was at a height of 17m by plugging in 17m for h:

$17=-5t^2 + 20t + 2$ .

Move the 17 to the other side, then solve by factoring / quadratic equation. You'll get two times. The ball was ABOVE 17m between those two times, so you can find the length of time simply by subtracting.

See how that goes!
June 19th 2008, 07:25 PM
Nikola

Thank you soo much! I have not tried it yet, im just about too! Thanks soo much! I will post my answer soon.

Edit: Something is wronge. Im not sure what though but ill start with the first one.
When i sub 0 it doesn't make a difference and when i do the quadratic formula i get -4 and 8 which is not the answer.
For the second one i subed 17. Here are my steps:

17 = -5t^2 + 20t + 2
0 = -5t^2 + 20t - 15

and i end up with -1 and 3! What am i doing wrong?
June 19th 2008, 07:33 PM
Nikola

Alright i've updated the thread since the new answer wont work! Please read previuse updated post.
June 19th 2008, 08:23 PM
Isomorphism

Quote:

Originally Posted by Nikola

I need to know if there is a standard method of finding the answer to this. My exam is tomorrow and my teacher gave us a set of topics to study. This is titled Projectile Motion!

The question is:

The function h=-5t^2 + 20t + 2 gives the approximate height, h meters of a trown football as a function of the time, t seconds since it was thrown. The ball hit the ground before the receiver could get near it.

a) How long was the ball in the air, to the nearest tenth of a second?
b) for how many seconds was the height of the ball at least 17 m?

The answers are:
a) 4.1
b) 2

I dont understand how to get the answers though. Sorry if there are misspelling in the question i had to rewrite it from the text book. Thanks again!

a) Well the ball was in air until it hit the ground. And when it hit the ground the height of the ball from the ground is 0.

$h=-5t^2 + 20t + 2 = 0 \Rightarrow t = \frac{-20 \pm \sqrt{20^2 - 4(-5)(2)}}{-10} =\frac{20 \pm 20.976}{10} = 4.0976$

I have ignored the negative root.

b) Read the question carefully. Its asking how long was the height of the ball greater than 17 meters. You have to find t such that

$-5t^2 + 20t + 2 \geq 17$

$\Rightarrow -5t^2 + 20t - 15 \geq 0$

$\Rightarrow t^2 - 4t + 3 \leq 0$

$\Rightarrow (t - 3)(t - 1) \leq 0$

$\Rightarrow t \in [1,3]$

So between 1 seconds and 3 seconds the ball was at a height greater than 17 meters. That means for 2 seconds the ball was at a height of at least 17 meters
June 19th 2008, 08:33 PM
Reckoner

Quote:

Originally Posted by Nikola

Thank you soo much! I have not tried it yet, im just about too! Thanks soo much! I will post my answer soon.

Edit: Something is wronge. Im not sure what though but ill start with the first one.
When i sub 0 it doesn't make a difference and when i do the quadratic formula i get -4 and 8 which is not the answer.
For the second one i subed 17. Here are my steps:

17 = -5t^2 + 20t + 2
0 = -5t^2 + 20t - 15

and i end up with -1 and 3! What am i doing wrong?

I think you are making a mistake when using the quadratic formula.

For $-5t^2 + 20t + 2 = 0$ , we have

$t = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a} = \frac{-20\pm\sqrt{440}}{-10}$

$t = 2\pm\frac{2\sqrt{110}}{-10} = 2\pm\frac{\sqrt{110}}{5}$

The negative root can be ignored since $t > 0$ . I don't see how you could have got -4 and 8.

The quadratic in the second part, $-5t^2 + 20t + 2 = 17$ , is easier since you can factor it over the integers, but you can use the quadratic formula if you like.

$-5t^2 + 20t + 2 = 17$

$\Rightarrow-5t^2 + 20t - 15 = 0$

$\Rightarrow t^2 - 4t + 3 = 0$

$\Rightarrow (t - 3)(t - 1) = 0$

$\Rightarrow t = 1\text{ or }t = 3$

So your time interval in seconds is $1\leq t\leq 3$ , so the ball was at or above 17 m for 3 - 1 = 2 seconds.
June 19th 2008, 08:42 PM
Nikola

Thanks guys! You where right i was doing my quadratic equation wronge (I include the square root in the calculation to save time) But just to make sure. Whenever i get a question like this i just factor or do a quadratic formula and I have my answer? Thanks again you've all been a great help!