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Projectile Motion homework problem

  • Thread starter Destrio
  • Start date
  • #1
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A projectile is fired from the surface of level ground at an angle Ao above the horizontal.
A) Show that the elevation angle B of the highest point as seem from the launch point is related to Ao by tanB = (1/2)tanAo
B) Calculate B for Ao = 45 deg

So
I figure at the highest point, Vy must equal 0.
let H be height of launch
let R by range of launch

so tanB = H/(1/2)R
and tanB = (1/2)tanAo

therefore
H/(1/2)R = (1/2)tanAo
tanAo = H/(1/4)R

I'm not sure how to proceed, do I want to derive H or R in terms of velocity/acceleration?

Thanks,
 

Answers and Replies

  • #2
learningphysics
Homework Helper
4,099
5
A projectile is fired from the surface of level ground at an angle Ao above the horizontal.
A) Show that the elevation angle B of the highest point as seem from the launch point is related to Ao by tanB = (1/2)tanAo
B) Calculate B for Ao = 45 deg

So
I figure at the highest point, Vy must equal 0.
let H be height of launch
let R by range of launch

so tanB = H/(1/2)R
and tanB = (1/2)tanAo
I'm confused... how did you do this last step? This is what you're supposed to prove, but you haven't proven it...

Find H and R in terms of Ao... then you can substitute them into tanB = H/(R/2)
 
  • #3
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I was just stating what it has to equal, I guess I can't do that since that is what I'm trying to prove

Can I use the statement: tanB = (1/2)tanAo to prove it?
for instance
H/(1/2)R = (1/2)tanAo
tanAo = H/(1/4)R

H = (1/4)R*tanAo
R = H/(1/4)tanAo
 
  • #4
learningphysics
Homework Helper
4,099
5
I was just stating what it has to equal, I guess I can't do that since that is what I'm trying to prove

Can I use the statement: tanB = (1/2)tanAo to prove it?
for instance
H/(1/2)R = (1/2)tanAo
tanAo = H/(1/4)R

H = (1/4)R*tanAo
R = H/(1/4)tanAo
no. you can't use the statement that you're trying to prove...

forget about B for now... try to get H and R in terms of Ao.... suppose you have an object with initial velocity v at an angle Ao... now what is the height in terms of v and Ao? What is the range in terms of v and Ao?
 
  • #5
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Alright

When Vy = 0
the height will = H
and the range will = (1/2)R

since the opposite and adjacent side of tanAo wont be in terms of H yet
let J = tanAo(1/2)R
since the adjacent side of both angles will be equal

for initial velocity

Vox = VocosAo
Voy = VosinAo

velocity components:
Vx = Vox
Vy = Voy - gt

for displacement
x = Voxt
y = Voyt - (1/2)gt^2

y = (tanAo)x -[g/(2VocosAo)]x^2

R = (2Vo^2)sinAocosAo / g
R = (Vo^2)(sin2Ao) / g
R = (2Vo^2)(cosAo^2)(tanAo) / g

tanAo = gR / (2Vo^2)(cosAo^2)

I'm getting lost here
 
  • #6
learningphysics
Homework Helper
4,099
5
I think you're making it more complicated than it needs to be...

You know the time taken to get to maximum height is Vosin(theta)/g. So the time to hit the range is 2Vosin(theta)/g

x = Vocos(theta)t

R = Vocos(theta)(2Vosin(theta)/g = 2Vo^2sin(theta)cos(theta)/g (this is what you got, but I'm not sure how you got it... I wasn't able to follow).

What is H?
 
  • #7
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is there a simpler way that you would approach R?

wont H just be tanAo x (1/2) R
except that it is on a curve

or H = Voyt - (1/2)gt^2

but at H Vy = 0
im sortof lost here
 
  • #8
Try this:

This is somewhat a special case
1 Vyf=Vyi+Ayt
2 0=Visin(yourangle)-gt1
3 t1=((Vi)/g)sin(yourangle)
4 h=[(Visin(yourangle)]∗((Visin(yourangle))/g)-(1/2)g(((Visin(yourangle))/g))²
5 h=((Vi²sin²(yourangle))/(2g))
6 R=Vxit2=(Vicos(yourangle)2t1
7 =(vicos(yourangle))((2Visin(yourangle))/g)=((2Vi²sin(yourangle)cos(yourangle))/g)
8 From trigonometry sin2(yourangle)=2sin(yourangle)cos(yourangle)
9 so R=((Vi²sin2(yourangle))/g)
10 range will max out at 45 degrees
 
  • #9
212
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"your angle" referring to the angle Ao (angle of the launch) as opposed to B (angle who's hypotenuse will touch the highest point)
 
  • #10
learningphysics
Homework Helper
4,099
5
is there a simpler way that you would approach R?

wont H just be tanAo x (1/2) R
no, H = tanB*(1/2)R

except that it is on a curve

or H = Voyt - (1/2)gt^2

but at H Vy = 0
im sortof lost here
plug in t = Vosin(theta)/g into

H = Vosin(theta)t - (1/2)gt^2

to get H...

or better yet.. using the kinematics equation vf^2 = vi^2 + 2ad. you know a = -g, vf = 0, vi = Vosin(theta)

what is d?
 
  • #11
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H = Vosin(theta)Vosin(theta)/g - (1/2)g(Vosin(theta)/g)^2
H = Vosin(theta)^2 / g - (1/2)Vosin(theta)^2 / g
H = Vosin(theta)^2 - (1/2)Vosin(theta)^2 /g
H = (1/2)Vosin(theta)^2 / g

or

vf^2 = vi^2 + 2ad
0 = Vosin(theta)^2 +2gd
d = -(Vosin(theta)^2) / 2(-g)
d = (1/2)Vosin(theta)^2 / g

can I use these to compare the 2 initial equations?
 
  • #12
learningphysics
Homework Helper
4,099
5
H = Vosin(theta)Vosin(theta)/g - (1/2)g(Vosin(theta)/g)^2
H = Vosin(theta)^2 / g - (1/2)Vosin(theta)^2 / g
H = Vosin(theta)^2 - (1/2)Vosin(theta)^2 /g
H = (1/2)Vosin(theta)^2 / g

or

vf^2 = vi^2 + 2ad
0 = Vosin(theta)^2 +2gd
d = -(Vosin(theta)^2) / 2(-g)
d = (1/2)Vosin(theta)^2 / g

can I use these to compare the 2 initial equations?
Now you know H and R...

R = 2Vo^2sin(theta)cos(theta)/g
H = (1/2)Vosin(theta)^2 / g


substitute them into tanB = H/(R/2)

simplify...
 
  • #13
212
0
awesome
thanks
 

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