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Projectile Motion- What angle does range equal max height?

  • Thread starter Acnhduy
  • Start date
  • #1
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1

Homework Statement



At what projection angle will the range of a projectile equal its maximum height?

Homework Equations



Big 5

vf2= vi2 + 2ad

The Attempt at a Solution



So I attempted to solve this question by randomly subbing in equations into each other.

If i draw out a diagram to solve for the angle, I would need the x and y components.

Let V equal the total velocity.

viy= Vsinθ
Plugging into vfy2= viy2 + 2aydy , I get

vf2= V2sin2θ + 2aydy

Now I isolate for d, and end up with dy= (V2sin2θ)/2ay

Ok, after I got the displacement vertically, since I am trying to find the point where they both are at equal magnitude, I now solve for displacement horizontally.

Now using the range formula, I can determine the horizontal displacement.

dx= (V2sin2θ)/g

(g=ay)
With these two formulas, I substitute and...

(V2sin2θ)/2ay = (V2sin2θ)/ay

sin2θ = aysin2θ

sin2θ = ay(2sinθcosθ)

sinθ = 2aycosθ

sinθ/cosθ = 2ay

tanθ = 2(-9.8)

θ = tan-1(-19.6)

θ = 87°

However... this is not the answer that the answer key states. The answer shown is 76°, and I do not know where I have gone wrong, or if I am doing this right.

Sorry if it was a mess and confusing, thanks for the help!
 

Answers and Replies

  • #2
6,054
390

Homework Statement



At what projection angle will the range of a projectile equal its maximum height?

Homework Equations



Big 5

vf2= vi2 + 2ad

The Attempt at a Solution



So I attempted to solve this question by randomly subbing in equations into each other.

If i draw out a diagram to solve for the angle, I would need the x and y components.

Let V equal the total velocity.

viy= Vsinθ
Plugging into vfy2= viy2 + 2aydy , I get

vf2= V2sin2θ + 2aydy

Now I isolate for d, and end up with dy= (V2sin2θ)/2ay
No. That should be $$ d_y = - \frac {v^2 \sin^2 \theta} {2a_y} $$ Note the minus sign.

Ok, after I got the displacement vertically, since I am trying to find the point where they both are at equal magnitude, I now solve for displacement horizontally.

Now using the range formula, I can determine the horizontal displacement.

dx= (V2sin2θ)/g

(g=ay)
With these two formulas, I substitute and...

(V2sin2θ)/2ay = (V2sin2θ)/ay
Assuming the above is correct (which it is, because you made another sign error there!), how do you get this:

sin2θ = aysin2θ
?
 
  • #3
31
1
I got sin2θ = aysin2θ

from

(V2sin2θ)/2ay = (V2sin2θ)/ay

Cross multiply and cancel out one of the ay

ay(V2sin2θ) = 2ay(V2sin2θ)

[STRIKE]ay[/STRIKE] (V2sin2θ) = [STRIKE]2 [/STRIKE]ay(V2sin2θ)

Divide both sides by V2

[STRIKE]V2[/STRIKE]sin2θ = ay([STRIKE]V2[/STRIKE]sin2θ)

sin2θ = aysin2θ
 
  • #4
6,054
390
You need to brush up on your algebraic manipulation skill. You do not cancel "one of the a" like that.
 
  • #5
31
1
If one side has "2a" and the other has "a", can't you divide out one of the "a" by dividing both sides by it?
 
  • #6
31
1
Ohhh
 
  • #7
6,054
390
If you divide both sides by "a", each side gets divided by "a". So if one side has "2a", and the has "a", how can the result of that be "a" and "1"? Do you really think that the inverse operation, "a" times "a", is "2a"?
 
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  • #8
31
1
Ok I see what you mean. I'm left with 2 instead of 'a'. Mistakened as a square
 
  • #9
6,054
390
So what do you get now?
 
  • #10
31
1
Thanks for the help :). After correcting that I end up with 76° :D
 

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