Projectile Problems: Range, Time & Theta C (6 Sig Digits)

[Taco John]

so does that mean that AJ's answers for d and e are incorrect at post 19? i have a similar problem and this is good practice.

 

[learningphysics]

Is theta critical arcsin[(2x2^(1/2))/3]?

Yes, that's the answer.

 

[Taco John]

another question...how are you using x numbers which are beyond the point of impact/where the projectile stops? wouldn't you have to use numbers below or equivalent to 115.021m?

 

[learningphysics]

another question...how are you using x numbers which are beyond the point of impact/where the projectile stops? wouldn't you have to use numbers below or equivalent to 115.021m?

The x is below the range... the y is below the height... however the distance from the start which is sqrt(x^2 +y^2) is above the max. height and range...

 

[Taco John]

so c is t = 5.55736s and d is 130.705m? trying to write these down...

and what do you use for the critical angle?

 

[Brentavo7]

Here are the answers I have

a. range of projectile: 115.020m
b. total time of flight: 9.70295s
c. time in which the projectile is farthest from its launch point: 5.55736s
d. farthest distance the projectile is from its launch point: 130.705m
e. the critical angle theta c such that for theta c is greater than or equal to theta which is greater than 0 the projectile is always moving farther away from its launch point and for theta greater than theta c there are times during the projectiles trajectory when the projectile is moving towards its launch point: 70.3144 degrees

Is this right guys?

 

[learningphysics]

The exact answer for the critical angle is as machinegoesping posted. arcsin(\frac{2\sqrt{2}}{3}) = 70.5288 degrees

 

[learningphysics]

Here are the answers I have

a. range of projectile: 115.020m

Looks good... but 115.02053... wouldn't this round to 115.021?

b. total time of flight: 9.70295s

I get 9.702957... which would round to 9.70296

c. time in which the projectile is farthest from its launch point: 5.55736s

correct.

d. farthest distance the projectile is from its launch point: 130.705m

correct.

e. the critical angle theta c such that for theta c is greater than or equal to theta which is greater than 0 the projectile is always moving farther away from its launch point and for theta greater than theta c there are times during the projectiles trajectory when the projectile is moving towards its launch point: 70.3144 degrees

Is this right guys?

see my previous post.

 

[Brentavo7]

really? what am I doing wrong. it seems simple enough.
ArcSin ((square root of 2 * 2)/3) = arcsin ((1.414214 * 2)/3) = arcsine (2.828427/3)
=Arcsine 0.942809 = 70.31.44 degrees according to my calculator.

 

[Brentavo7]

Good catch on the rounding. Thanks. now if I can just figure out my mistake on the angle I am all set.

 

[Taco John]

^ i noticed that too...and my last post on this :) I'm stealing some thunder from brentavo :)

what formula were you using for the farthest distance (part d)? thanks much...now i hope my teacher asks something like this for HW :)

 

[learningphysics]

really? what am I doing wrong. it seems simple enough.
ArcSin ((square root of 2 * 2)/3) = arcsin ((1.414214 * 2)/3) = arcsine (2.828427/3)
=Arcsine 0.942809 = 70.31.44 degrees according to my calculator.

Hmmm... strange. maybe my calculator is messed up. :( I'm plugging in 0.942809 and taking arcsin... 70.5288... do you have another calculator to check? maybe it's just my calc...

 

[learningphysics]

^ i noticed that too...and my last post on this :) I'm stealing some thunder from brentavo :)

what formula were you using for the farthest distance (part d)? thanks much...now i hope my teacher asks something like this for HW :)

Take the derivative of this equation:

d^2 = x^2 + y^2

2d*dd/dt = 2x*dx/dt + 2y*dy/dt

set dd/dt = 0

x*dx/dt + y*dy/dt = 0

plug in the stuff... divide by t and you get a quadratic.

 

[Taco John]

no, your numbers are right, learning. he might have it in radian mode or something...one of those silly things.

 

[Brentavo7]

Looks like it was mine. glad I got the other answers right with it. Thanks for all the help I appreciate it.

 

[learningphysics]

no, your numbers are right, learning. he might have it in radian mode or something...one of those silly things.

Thanks Taco John.

 

[learningphysics]

Looks like it was mine. glad I got the other answers right with it. Thanks for all the help I appreciate it.

no prob.

 

[Taco John]

ok, i lied...wasn't the last post (so i lied twice i guess). lol.

what am i plugging in? i have x = 11.8542m/s and y = 47.5445m/s...those the x and y numbers i use? or am i on the wrong track entirely?

 

[Brentavo7]

use the formula I had in post 21 for question c. your Vox and Voy are correct. I used this site to cheat. http://www.csgnetwork.com/quadraticcomplex.html. when you get the time just figure out your x and y at that time. AJ had it earlier on post 22. then use x^2 + y^2 = d^2 to get your distance at that point.

 

[Brentavo7]

the projectile is 128.87m away from the launch point at the half way point. But 130.705m away at the max.

 

[Taco John]

you just completely lost me...what are you using that site for? like what are you using for the quadratic? i never got that far (i dislike derivatives like crazy)

 

[Taco John]

nm, I'm getting tired...i got it...thanks :)

 

[Brentavo7]

yeah I am not too fond of them either. I took this equation
(Vox*t)(Vox) + (Voy*t-.5*g*t^2)(Voy-g*t) = 0 divide both sides by t you get
Vox^2 + (Voy - .5*g*t)(Voy - g*t) = 0 you can create a quadratic with this
Vox^2 + Voy^2 - (.5*g*t*Voy) - (g*t*Voy) + (.5*g^2*t^2)
plug in g= 9.8; Vox= 11.8542; Voy= 47.5445 youll get a quadratic that you can plug into the site I linked

 

[Brentavo7]

(Vox^2 + Voy^2) - (14.7t*Voy) + (48.02t^2) = 0
2400.9998 - 698.904t + 48.02t^2 = 0