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

[Brentavo7]

Homework Statement

projectile intial velocity vo of 49 meters per second. angle theta intial is 76 degrees. projectile takes off and lands on the same horizontal plane. neglecting air resistance and using g=9.8 m/s squared. find

a. range of projectile: 115.020m
b. total time of flight: 9.70295
c. time in which the projectile is farthest from its launch point: 4.85147
d. farthest distance the projectile is from its launch point: 115.330
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: not sure where to begin

give six significant digits in all answers

Homework Equations

Vox = Vo Cos(theta); Voy = Vo sin(theta); 1/2t=Voy/ay; t=2(1/2t)
dy=Voyt+.5ayt^2; dx=Voxt

The Attempt at a Solution

Vox= 49 cos76= 11.85417
Voy= 49 sin76= 47.54449
1/2t= 47.54449/9.8= 4.851479
dy= (47.54449)(4.851479)+(.5)(-9.8)(4.851479)(4.851479)=230.6611-115.3306= 115.330
t= 2(4.851479)= 9.70295
dx= (11.85417)(9.702958)= 115.020


I have been out of school for quite sometime now and didn't do so well on physics then. I believe I am close on the range and total time. I feel I am missing something for farthest point. I was assuming this meant the peak of its arc from the ground. But after reading it again I am not so sure. And Theta C I don't get since its an arc not an angle unless its talking about the angle of launch. Any hint in the right direction would be appreciated. Thanks.

 

[learningphysics]

EDIT: Sorry! thanks chocokat! silly of me... because dy is just 0 at the range anyway.

Looks like all your answers are correct Brentavo7!

 

[chocokat]

Learningphysics - since the projectile goes up and comes back down, at 9.7$# it'll be back on the ground (or at the same vertical place where it started), whereas, won't using the 4.85$# it'll calculate the top of the arc, i.e. the farthest distance from the launch point?

 

[learningphysics]

Learningphysics - since the projectile goes up and comes back down, at 9.7$# it'll be back on the ground (or at the same vertical place where it started), whereas, won't using the 4.85$# it'll calculate the top of the arc, i.e. the farthest distance from the launch point?

Yes, you're absolutely right. Thanks!

 

[Saladsamurai]

So at a projectile's max height, it is furthest from its launch point? (I am not in doubt that you are correct, I am just having trouble reasoning why).

Casey

 

[Dick]

So at a projectile's max height, it is furthest from its launch point? (I am not in doubt that you are correct, I am just having trouble reasoning why).

Casey

I am in doubt that's correct, because I don't think it is. And I have no clue what ### means.

 

[chocokat]

No, the furthest point may or may not be the height, but in this case it is (based on my calculations, which I hope are correct), but just by a little.

Height = 115.33m
Horizontal distance = 115.02m

 

[chocokat]

Dick - sorry, I used ### as I only went out 3 places, but the original question needed 6, and I didn't bother to get the rest of the places.

 

[learningphysics]

So at a projectile's max height, it is furthest from its launch point? (I am not in doubt that you are correct, I am just having trouble reasoning why).

Casey

Yes, you're right. I think I'll avoid posting on this thread for a little while... think I'm a little too tired tonight. :tongue:

 

[Dick]

Dick - sorry, I used ### as I only went out 3 places, but the original question needed 6, and I didn't bother to get the rest of the places.

That's ok. But at the farthest distance from the launch point the velocity should be perpendicular to a line joining it to the launch point. Can't be true at the top point of the trajectory if there is any horizontal velocity. Significant places are another matter and you are right that the difference is smallish. But the OP was asked for six sig figs.

 

[learningphysics]

So parts a and b are good. for part c)

distance from the launch point d^2 = x^2 + y^2

take the derivative with respect to time

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

since dd/dt = 0 at the max...

0 = x*dx/dt + y*dy/dt (giving that (x,y) is perpendicular to (dx/dt, dy/dt) as Dick mentioned)

plug these in and solve for t... I think it will slightly more than 4.851479...

then for part d), get x and y at the time you found in part c).

For part e)... if the object is always moving away from the launch point... that means dd/dt always > 0

ie x*dx/dt + y*dy/dt always >0... following Dick's suggestion from another thread, take equation x*dx/dt + y*dy/dt = 0... this reduces to a quadratic... find the range of the angle theta such that this has no real roots.

 

[Brentavo7]

Thank you all for the help. I now see that the height was not the farthest from the launch point. Using the formulas mentioned. I am now getting the farthest distance of130.055 meters at 5.38863 seconds. Critical Angle of 71.4856 degrees. Still not positive I am doing the angle correctly. I've got to get some sleep though. Ill try again in the morning. Thanks again.

 

[learningphysics]

Thank you all for the help. I now see that the height was not the farthest from the launch point. Using the formulas mentioned. I am now getting the farthest distance of130.055 meters at 5.38863 seconds. Critical Angle of 71.4856 degrees. Still not positive I am doing the angle correctly. I've got to get some sleep though. Ill try again in the morning. Thanks again.

I'm not getting exactly the same answers (but close though). Can you show how you got them?

 

[Brentavo7]

I actually ended up cheating a bit. I used autocad to map out the curve and placed a circle from the launch point to the max height. saw where the curve intersected the circle again. Took the mid point of the two intersections as the farthest point from the launch point. This gave me a distance of 130.076m at 5.33971s. I would like to verify by formula. I am confused on dd/dt=0.

 

[learningphysics]

I actually ended up cheating a bit. I used autocad to map out the curve and placed a circle from the launch point to the max height. saw where the curve intersected the circle again. Took the mid point of the two intersections as the farthest point from the launch point. This gave me a distance of 130.076m at 5.33971s. I would like to verify by formula. I am confused on dd/dt=0.

Oh I see... I think that's pretty cool that you were able to use autocad for that.

The algebra is a bit cumbersome for the problem...

starting here:
d^2 = x^2 + y^2

d is the distance of the projectile from the start point.

you want to find the spot where dd/dt = 0. Generally the max or min of a variable occurs when its derivative is 0.

so the first step is taking the derivative with respect to t, of

d^2 = x^2 + y^2

 

[Brentavo7]

ok I am beginning to grasp it. I seem to be reverse engineering though. if i use the formula
d(dd/dt) = x(dx/dt) + y(dy/dt) and plug in the numbers I found. I get
130.0765 * 130.0765 /5.33971 = (63.29792 * 63.29792 /5.33971)+
(113.6366 + 113.6366 /5.33971)

16919.9 / 5.33971 = (4006.627 / 5.33971) + (12913.28 / 5.33971)
3168.693 = 750.3454 + 2418.349
3168.693 = 3168.694

As far as angle Theta c is concerned am I closer with the original 71 something degrees or is it more common sense of 67.5

 

[Brentavo7]

Thanks for the compliment. I've been a fire protection engineer for 6 years now. We use Autocad extensively. I've got Angle Theta C to be somewhere between 71.85 and 71.9 degrees. Does anyone happen to know the number out to six significant figures.

 

[learningphysics]

I recommend going through the algebra here... it's actually not too bad now that I go through it... although your numbers are close, they aren't close enough...

x*dx/dt + ydy/dt = 0

$$(vcos(\theta)t)(vcos\theta) + (vsin(\theta)t - (1/2)gt^2)(vsin(\theta)-gt) = 0$$

I recommend going through it without plugging in numbers till the very end.

first thing you can do is divide by t...

you'll get a quadratic in terms of t...

use the quadratic formula to solve for t...

so this actually takes care of both the time to get the maximum distance... and also the angle... when the discriminant< 0 there is no solution... and that gives the critical angle.

I'll verify your expression when you get it.

 

[Ajwrighter]

A)

R = Vo2 Sin(2θ)
g​

R = (49m/s)2 Sin(2(76o))
9.8m/s2
R = 115.02053288253m​

B)

TFinal = 2 Vo Sin(θ)
g​

TFinal = 2 (49m/s)(Sin76o)
9.8
TFinal = 9.702957263s​

C)

Y = -(1/2)(g)(t2) +Vo Sin(θo)t
= -4.9 t2 + 47.544490587524t – 115.33054006261

then t => -b +/- √((b2) – (4)(a)(c))
2a
t = 4.85147863138s​

D)

H = Vo2 Sin2 (θ)
2g

= (49m/s) 2 Sin2 (76)
2(9.8)

H = 115.33054006261m​

E)

X = Vo Cos(θ)t
X = 57.510266441273m

Y = -(1/2)(g)(t2) +Vo Sin(θo)t
Y= 115.33054006261m


θc =ArcTan(Y/X)

θc = 63.496586096704o​

Anyone want to check that? I think that’s right.

 

[Ajwrighter]

some of the ^2 came out as just 2 but the answeres are still correct.

 

[Brentavo7]

Hey Andy check c and d. The height isn't the farthest point. use the equation in post 18.

My math may be wrong but I am getting:

(Vox^2 + Voy^2) - (14.7t*Voy) + (48.02t^2) = 0
2400.9998 - 698.904t + 48.02t^2 = 0
Im getting t = 5.55736. That right?

 

[Ajwrighter]

hmm at 5.55736s wouldn't that place the object at (x,y) (65.87790622,112.8890244) versus (57.510266441273, 115.33054006261m) hmm I will go back over it later tonight. to tired to try and do any more math right now lol.

 

[Brentavo7]

The math seems to work for 5.55736 but I also get 8.99707 out of the quadratic.

 

[Ajwrighter]

use my numbers and see if you get the same answere.

 

[Brentavo7]

with t= 4.85147 the distance from the launch point is 128.87m using x^2 + y^2 = d^2
at t= 5.55736 the distance from the launch point is 130.705m

 

[learningphysics]

with t= 4.85147 the distance from the launch point is 128.87m using x^2 + y^2 = d^2
at t= 5.55736 the distance from the launch point is 130.705m

exactly the same numbers I got. :) did you get the critical angle?

 

[Brentavo7]

nope still confused on the angle.

 

[learningphysics]

nope still confused on the angle.

Solve for t, without plugging in any numbers (do the same thing you did before, but without plugging in numbers, just use the variables)... use the quadratic formula... when the discriminant is less than 0, this has no solution... ie dd/dt never hits 0... hence dd/dt always remains above 0...

when you set the discriminant less than 0, you get the critical angle.

 

[machinegoesping]

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

 

[Brentavo7]

Vox^2 + (Voy2 - 14.7*Voy + 48.02)
Im getting confused with the sine and cosine of the angles

 

[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?