1. ## Quadratic Formula Question #6

While stopped at the Port of Barcelona on your European cruise, you decide to try some skeet shooting. Skeet shooting is an activity where clay discs are flung through the air by a machine. The objective is to shoot and hit them before they fall to the ground. The path of the clay disc can be represented with the parabolic function:

h(t)
= -4.9t2 + 32t + 2

where h(t) is the height in metres after t seconds.

a) Determine the initial height of the clay disc as it leaves the skeet shooting machine.

My attempt: So, I'm not sure how I can solve for height, should I make t = 0?

If I did that, wouldn't that make the equation turn into h = -4.9 + 32 + 2? If I solved for h like that, would that be correct? I'm not sure...

b) Determine the amount of time it takes for the clay disc to hit the ground if it does not get shot. Round your answer to the nearest tenth.

My attempt: h(t) = -4.9t2 + 32t + 2

0 = -4.9t2 + 32t + 2

Used quadratic formula... t = 6.59 s Is this correct?

c) Determine the maximum height of the clay disc and the amount of time it takes to reach that height. Round both answers to the nearest tenth.

h(t) = -4.9t2 + 32t + 2

My attempt: So, I'm not to sure about this one either, I graphed this equation into my calculator, and would the maximum be the time if took for the disk to reach the max height, or is it the max height?

d) What is the height of the clay disc after 4 seconds? Express your answer to the nearest tenth of a metre.

My attempt: I went into the table in my calculator and went to x = 4 and it says Y1 = 51.6 , is the answer to this question 51.6 m high?

e) If the clay disc travels at 60 km/ h, determine how many metres it will travel when it hits the ground. Round your answer to the nearest tenth of a metre.

My attempt: h(t) = -4.9t2 + 32t + 2

When the disk is at the ground, h = 0

0 = -4.9t2 + 32t + 2

Solve for t... = 6.47

I don't know what next...

f) If you hit the clay disc at a height of 20 metres, then how much time passed from the time it was released until it was hit? Find both possibilities. Round your answers to the nearest tenth.

My Attempt: (20)t= -4.9t2 + 32t + 2

0 = -4.9t2 + 12t + 2

Use quadratic formula... d = 2.6 m??? Is this correct?

g) When Bryan tries to hit the clay disc, the path of the pellet that hits the disc can be modelled by the equation h(t) = 5t. At what height does Bryan’s pellet hit the disc? Round your answer to the nearest tenth of a metre.

I do not know what to do...

Any help would be great! Thanks in advance!

2. ## Re: Quadratic Formula Question #6

Hey Mathnood768.

For initial height, you are correct in that it corresponds to t = 0. You have the right idea for b) [I haven't checked the answer though].

For c) you need to find where the height is a maximum and for this you use calculus. Have you covered derivatives?

For d) it is correct and remember that positive values mean values above the ground whereas negative values mean values below the ground (but in your problem you don't get negative values).

Hint for e) distance = velocity * time (if velocity is constant).

For f) Remember that h(t) = 20 = -4.9t^2 + 12t + 2 so -4.9*t^2+ 12*t - 18 = 0.

Hint for g) When they intersect it means that h(t) = g(t) where h(t) is equation for pellet height and g(t) is equation for clay disc height both as a function of time. Try re-arranging to solve for t (you should end up with a quadratic equation).

3. ## Re: Quadratic Formula Question #6

For (c) . we know that the height at t=0 is 2. Since the skeet is going up and coming down, we know that the time for a body for going up is equal ti time for coming down.
With this premise we will try to see when the skeet is going to be at height 2 That is the value of t when h(t) = 2
We have -4.9t^2 + 32 t + 2 = 2 OR t( - 4.9 t + 32 ) = 0
That gives t = 0 and t = 320/49. In this time the skeet has gone up and come down. So the time for going up or reaching maximum height will be 320/98.
Find h(t) for this value of t add 2 and that will give the maximum height.

4. ## Re: Quadratic Formula Question #6

Originally Posted by chiro
...

For c) you need to find where the height is a maximum and for this you use calculus. Have you covered derivatives?
For quadratic functions, one may simply locate the axis of symmetry, without having to use differential calculus.

For a quadratic given as:

$f(x)=ax^2+bx+c$

The axis of symmetry is the line:

$x=-\frac{b}{2a}$

and this is the line along which the vertex will be found, and the vertex is the global extremum.

5. ## Re: Quadratic Formula Question #6

What about the other questions?

6. ## Re: Quadratic Formula Question #6

Since when is \displaystyle \begin{align*} -4.9 (0)^2 = -4.9 \end{align*} and \displaystyle \begin{align*} 32(0) = 32 \end{align*}?

7. ## Re: Quadratic Formula Question #6

a) b) ??? I'm still confused, I showed my work, can someone tell me where I went wrong and help me?

8. ## Re: Quadratic Formula Question #6

Did you reconsider a) given Prove It's question above?

9. ## Re: Quadratic Formula Question #6

I don't understand, I'm not good at math, I am looking for someone to explain to me nicely where I went wrong and how I should go about this, because I still don't know. I get it t = 0, that was my only attempt, I didn't know what else to do.

10. ## Re: Quadratic Formula Question #6

Your calculations were wrong! You are correct that you need to let t = 0, but when you multiply numbers by 0, you do NOT get the same numbers, you should get 0. Fix it!

11. ## Re: Quadratic Formula Question #6

for c) is the max height 54.24 m? I graphed it and got the maximum.

12. ## Re: Quadratic Formula Question #6

so, if t = 0 everything multiplied with t = 0? the initial height = 2?

how about b) ?

13. ## Re: Quadratic Formula Question #6

I suggest working on ONE question at a time. You haven't even got the answer to part a) yet, so why not try to get that finished first? You are correct that you have to let t = 0, and you have substituted correctly. But your simplification is wrong, that is what you need to fix.

People get annoyed when you get given some instructions and you don't follow them, instead just complaining about it being so hard and wanting someone else to do it for you. You will get very little help that way...

14. ## Re: Quadratic Formula Question #6

Do you not understand what "h(t)= -4.9t^2+ 32t+ 2[/tex]" means? You are told that "h(t) is the height in metres after t seconds" so, yes, h(0), the height after 0 seconds- the initial time. Now, what is $h(0)= -4.9(0)^2+ 32(0)+ 2$. What do you get when you multiply 32 by 0? What do you get when you multiply -4.9 times 0 squared?

15. ## Re: Quadratic Formula Question #6

i guess h = 2! still not sure about the rest, for d) i tried making all the t = 4 but i got h = 514 m, i think that is too much...

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