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

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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.
 
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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?
 
Taco John said:
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...
 
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?
 
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?
 
Brentavo7 said:
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.
 
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.
 
Good catch on the rounding. Thanks. now if I can just figure out my mistake on the angle I am all set.
 
^ 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 :)
 
Brentavo7 said:
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...
 
Taco John said:
^ 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.
 
no, your numbers are right, learning. he might have it in radian mode or something...one of those silly things.
 
Looks like it was mine. glad I got the other answers right with it. Thanks for all the help I appreciate it.
 
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?
 
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.
 
the projectile is 128.87m away from the launch point at the half way point. But 130.705m away at the max.
 
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)
 
nm, I'm getting tired...i got it...thanks :)
 
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
 
(Vox^2 + Voy^2) - (14.7t*Voy) + (48.02t^2) = 0
2400.9998 - 698.904t + 48.02t^2 = 0
 

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