Question

Precalc word problem, please help? The range of a projectile that is launched with an initial velocity v at an angle of...

Answer 1

theta is given by \[R=v^2/g \sin2\theta\]

Answer 2

where g is the acceleration due to gravity or 9.8 meters per second squared. If a projectile is launched with an initial velocity of 15 meters per second, what angle is required to achieve a range of 20 meters?

Answer 3

\[R=(v^2/g)\sin2 \theta\]

Answer 4

Yes, but I get to \[\sin2 \theta=17.66\] and don't know what to do from there.

Answer 5

Oh wait that's wrong

Answer 6

Sorry for my wrong typo. Solve \[\sin 2 \theta =\frac{ 20 \times 9.8 }{ 15^2 }\]

Answer 7

You should get 30.29

Answer 8

as the answer

Answer 9

So, \[\sin2 \theta=0.87\] How do I get from there to 30.29?

Answer 10

Take the arc sine of 0.87 and then divide the answer by 2

Answer 11

that is \[\sin^{-1} (0.87) \over 2\]

Answer 12

Ohh, okay that makes sense. But my answer key says there is a second answer, 59.71?

Answer 13

@girlnotonfire I think you mixed up something the actual formula is \[R=\frac{ v^{2}\sin 2\theta }{ g }\]

Answer 14

30.29+59.71 = 90, rings a bell

Answer 15

@azetina good one.

Answer 16

The formula simplifies to \[R=\frac{ v^{2} }{ g }\] when theta is 90 degrees.

Answer 17

SORRY NOT THETA BUT 2 TIMES THETA.

Answer 18

I still don't completely understand, but thanks guys.

Answer 19

.. @girlnotonfire can you post a quick screenshot of the material?

Answer 20

do you know how to take a screenshot?

Answer 21

Yes @jdoe0001 just give me a sec

Answer 22

so R = range

an the Range equation is given, so if we want to achieve 20 meters of Range, set R = 20 and solve by \(\theta\) that is \(\bf Range = R = \cfrac{v^2}{g}sin(2\theta)\qquad g=9.8\qquad v=15\qquad R=20\\ \quad \\
R = \cfrac{v^2}{g}sin(2\theta)\implies 20= \cfrac{15^2}{9.88}sin(2\theta)\)

Answer 23

woops... 9.8 not 9.88 dohh ehhe \(\bf Range = R = \cfrac{v^2}{g}sin(2\theta)\qquad g=9.8\qquad v=15\qquad R=20\\ \quad \\
R = \cfrac{v^2}{g}sin(2\theta)\implies 20= \cfrac{15^2}{9.8}sin(2\theta)\)

Answer 24

Okay, that's what I did first, but I was confused with what to do when I got to \[\sin2 \theta=0.87\]

Answer 25

so... recall that \(\bf somefunction^{-1}(somefunction(\theta))=\theta\)

thus \(\bf sin(2\theta)\approx 0.87\qquad \textit{taking }sin^{-1}\textit{ to both sides}\\ \quad \\
sin^{-1}[sin(2\theta)]\approx sin^{-1}(0.87)\implies \theta \approx sin^{-1}(0.87)\)

Answer 26

woops even forgot the 2... shoot anyhow

\(\bf sin(2\theta)\approx 0.87\qquad \textit{taking }sin^{-1}\textit{ to both sides}\\ \quad \\
sin^{-1}[sin(2\theta)]\approx sin^{-1}(0.87)\implies 2\theta \approx sin^{-1}(0.87)\)

Answer 27

Okay so \[\sin^{-1}(0.87)/2 = \theta \]

Answer 28

yeap

Answer 29

which is around 30 degrees

Answer 30

My answer key gives me a second answer (59.71), which @azetina pointed out was 90-30.29. Is that the way I'm supposed to get that second answer or...?

Answer 31

when you take arc sine, remember both theta and (180-theta) have the same sine value.

Answer 32

hmm I can see where the other is added.... I ... but 59.71 would be a complementary angle,

Answer 33

arc sine (0.87) = 60.46 or 119.54
theta = 1/2 * arc sine (0.87) = 60.46/2 or 119.54/2

Answer 34

because sin(x) and sin(180-x) have the same value.

Answer 35

hmm well.. I can see that....

I mean @girlnotonfire keep in mind that inverse sine has constraints of only the 1st and 4th Quadrants, so the angle will show up there when using that function

but @ranga is pointing out that, of course, reference angles on other Quadrants, or even cycles, will have the same sine

Answer 36

I understand that a positive sine, will give me an angle on the 2nd quadrant.... but hmm I just don't get a projectile launched at that angle =)

Answer 37

The actual angle is half of arc sine. So about 30 and 60 degrees angle launch are okay.

Answer 38

I kinda understand... Thanks @jdoe0001 you've been really helpful!!

Answer 39

yeap.. around 30 degrees