Question

Help with horizontal distance, initial velocity and angles problem? Medal and fan
17. Recall that the equation for the horizontal distance h in feet of a projectile with initial velocity v0 and initial angle theta is given by h=v0^2/16 sin theta cos theta.



a. Assume the initial velocity is 60 feet/second . What initial angle will you need to ensure that the horizontal distance will be exactly 100 feet? (3 points)

b. Assume the initial velocity is 60 feet/second. What is the maximum horizontal distance possible and at what angle does this occur?

Answer 1

Could you help me with this one? @zepdrix

Answer 2

I'm not familiar with that equation.
Are the sine and cosine in the denominator or no? :d

Answer 3

No, they aren't. They're next to it.

Answer 4

\[\Large\rm h=\frac{v_{o}^2}{16}\sin \theta \cos \theta\]k

Answer 5

For part a,
we let \(\Large\rm v_{o}=60\) and \(\Large\rm h=100\) and try to solve for our angle \(\Large\rm \theta\), yes?

Answer 6

Yes.

100 = (60)^2/ 16 sin theta cos theta

Answer 7

Remember your Sine Double Angle Identity? :)

Answer 8

sin(2x) = 2sinxcosx?

Answer 9

\[\Large\rm 100=\frac{(60)^2}{16}\color{orangered}{\sin \theta \cos \theta}\]Ok good.
So we need a 2 right?
Hmmm where can we get a 2 from?

Answer 10

Just multiply both sides by 2?

Answer 11

Yah there we go :3\[\Large\rm 200=\frac{(60)^2}{16}\color{orangered}{2\sin \theta \cos \theta}\]

Answer 12

\[\Large\rm 200=\frac{(60)^2}{16}\color{orangered}{\sin(2\theta)}\]

Answer 13

Wouldn't you multiply the (60)^2/16 as well?

Answer 14

Multiply that to the other side?
Yah seems like a good idea.

Answer 15

\[\Large\rm \frac{200\cdot 16}{60^2}=\color{orangered}{\sin(2\theta)}\]

Answer 16

Ah sorry were you typing it? >.< Did I steal the fun? lol

Answer 17

I gotta learn to slow down sometimes :P heh

Answer 18

Noo, I meant multiplying it by the 2.

Answer 19

Oh oh oh.
No, we only multiply the right side by 2 `one time`.

Recall multiplication:\[\Large\rm 2(3x)\quad \ne\quad 2\cdot3\cdot 2\cdot x\]We don't give a 2 to each in this example.

Answer 20

Don't confuse multiplication with exponents!
We certainly do ... do that with exponents.\[\Large\rm (3x)^2\quad=\quad 3^2\cdot x^2\]

Answer 21

I gave a 2 to the right side already.
See how I put it next to the sin theta?

Answer 22

Or did you mean something else +_+

Answer 23

Noo, you're right. Sorry I'm tired.

Answer 24

Okay, I gotcha so far.

Answer 25

So far we have:\[\Large\rm number=\sin(2\theta)\]

Answer 26

Understand how we can solve for theta from here?
Remember how to use your inverse sine function and stuff?

Answer 27

Sorry, I'm trying to remember currently. That was a bit ago.

Answer 28

Wouldn't it be x= 1/2 sin^-1y?

Answer 29

\[\Large\rm stuff=\sin(\text{angle)}\]Remember with inverse functions we swap the x and y?
Inverse trig functions are the same idea,
we swap the stuff with the angle.\[\Large\rm \text{angle}=\sin^{-1}(stuff)\]

Answer 30

Okay...was my answer right?

Answer 31

I dunno I'm not sure what you were doing >.<
Where is the 1/2 coming from?
Is that from the 2 that was connected to theta?

Answer 32

Yeah, it was. I'm sorry, I can't remember that at all for some reason. I don't understand how you can have arcsine 2 theta when it's not even in the range..

Answer 33

No that's fine :)
I guess you were just moving ahead.
I got confused because you changed the letters to x and y.\[\Large\rm number=\sin(2\theta)\]Applying the inverse,\[\Large\rm 2\theta=\sin^{-1}(number)\]Dividing by 2,\[\Large\rm \theta=\frac{1}{2}\sin^{-1}(number)\]

Answer 34

I got a little lazy with our big messy number...
let's see it was ummm...\[\Large\rm number=\frac{200\cdot16}{60^2}\]

Answer 35

So we need a calculator at this point :o

Answer 36

Yeahhh, that's where I was going, sorry!

Answer 37

Okay, hold on. Are we multiplying 1/2 arcsin by 2 then by sin again?

Answer 38

Whut? :U
No. We're not involving sine anymore.
Only arcsine.

Answer 39

So angle 30...are we plugging that back into the original sinthetacostheta?

Answer 40

That answers part a.

Answer 41

So we're working on b now? The answer to A was theta = 30?

Answer 42

Umm I came up with 31.4 degrees.

Answer 43

Wait for what?? Cause arcsin 1.2 is 30 degrees.

Answer 44

1/2*

Answer 45

\(\Large\rm \dfrac{200\cdot16}{60^2}\approx 0.888889\)

\[\Large\rm \frac{1}{2}\arcsin(0.888889)\approx 31.36697778\]

Answer 46

The 1/2 is OUTSIDE of the arcsine. :o
we're taking the arcsine of our big messy number.

Answer 47

And then multiplying that result by 1/2

Answer 48

Get confused in there somewhere? :o

Answer 49

Yeah haha, now I understand why we used the inverse instead of regular sin. Okay, so the number that is achieved for theta is 31.4?

Answer 50

Mmmm yah looks that way >.<

Answer 51

Alright awesome. So for b, we don't have h=100 and our angles aren't given again..

Answer 52

Part b is a little trickier.\[\Large\rm h=\frac{60^2}{16}\sin \theta \cos \theta\]We want to maximize our h.

Answer 53

How would we maximize it?

Answer 54

I'm gonna do something a little sneaky so we can write this as a single trig function again.
Instead of multiplying each side by 2,
I'm going to multiply the right side by 2 and 1/2 ok?\[\Large\rm h=\frac{1}{2}\cdot\frac{60^2}{16}\cdot2\sin \theta \cos \theta\]

Answer 55

We need that 2 for our Sine Double Angle Identity.
The 2 in the denominator we'll just combine with the 16.\[\Large\rm h=\frac{60^2}{32}\color{orangered}{2\sin \theta \cos \theta}\]

Answer 56

Was that step ok?
I was essentially giving the right side a 2/2 and placing each 2 in a different spot.

Answer 57

Yeah that makes sense!

Answer 58

\[\Large\rm h=\frac{60^2}{32}\color{orangered}{\sin2\theta}\]

Answer 59

Right on, so we'll use the same 1/2 arcsin?

Answer 60

No no no.
We would use the inverse function if we needed an `angle`.
Here we need height.
So we need to do something different.

Answer 61

Do you remember what the `amplitude` of a sine function tells us?\[\Large\rm y=A \sin x\]
\(\Large\rm A=amplitude\)

Answer 62

Yeah..would that mean that the big number is the amplitude?

Answer 63

The amplitude tells us how `big` our function can get: and max and min values.
Yes, that coefficient in front of our sine is our amplitude.

Answer 64

That number will be our maximum height.
We need to figure out `where` it happens though.

Answer 65

\(\Large\rm h=112.5\color{orangered}{\sin2\theta}\)

So we have something like this, yes?
And we want to know what `angle` gives us our maximum, 112.5.
In other words, let h=112.5 and solve for our angle again.

Answer 66

Ohhhh, alright. So would we plug in 112.5 into 1/2 arcsin?

Answer 67

No.
You're plugging 112.5 into h.

Answer 68

\[\Large\rm 112.5=112.5\sin 2 \theta\]

Answer 69

Ohhhhhhhh, then you would divide and the left side would be 1?

Answer 70

Mmm good!\[\Large\rm 1=\sin 2\theta\]

Answer 71

So the angle is 45?

Answer 72

yay good job \c:/

Answer 73

Awesome, thanks so much(: Sorry I was so slow, I'm working on my AP Chem assignment at the same time. I somehow turned out to excel at chemistry but definitely not math lol

Answer 74

hehe XD