# Projectile Motion- What angle does range equal max height?

• Acnhduy
In summary, the conversation discusses a question about finding the projection angle at which the range of a projectile equals its maximum height. The person attempted to solve the question using various equations and a diagram. However, they made a sign error in their calculations which led to an incorrect answer. After correcting the error, they were able to correctly determine the angle to be 76°.
Acnhduy

## Homework Statement

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

Big 5

## 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!

Acnhduy said:

## Homework Statement

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

Big 5

## 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

?

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

You need to brush up on your algebraic manipulation skill. You do not cancel "one of the a" like that.

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?

Ohhh

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"?

1 person
Ok I see what you mean. I'm left with 2 instead of 'a'. Mistakened as a square

So what do you get now?

Thanks for the help :). After correcting that I end up with 76° :D

## 1. What is projectile motion?

Projectile motion is the motion of an object through the air or space, under the influence of gravity. This type of motion follows a curved path, known as a parabola.

## 2. What factors affect the trajectory of a projectile?

The factors that affect the trajectory of a projectile include the initial velocity, launch angle, and the force of gravity. The air resistance can also have a small impact on the trajectory.

## 3. How do you find the maximum height of a projectile?

The maximum height of a projectile can be found by using the formula h = (v02sin2θ)/2g, where v0 is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.

## 4. What angle will give the maximum range for a projectile?

The angle that will give the maximum range for a projectile is 45 degrees. This angle provides the most balance between vertical and horizontal velocity components, resulting in the longest distance traveled.

## 5. Why does the angle for maximum range also equal the angle for maximum height?

This is because the maximum height and maximum range occur at the same time during the projectile's motion. At the angle of 45 degrees, the vertical and horizontal components of the initial velocity are equal, resulting in the maximum height and range being achieved simultaneously.

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