Project Motion/Trigonometry Question

[MrDickinson]

Homework Statement

A projectile is fired in such a way that its horizontal range is equal to three times its maximum height. What is the angle of Projection.

Relevant Equations

There are no given equations for this problem.

My reasoning and answer is wrong, but I cannot figure out why.

Perhaps it is strange, perhaps not, but I want to figure out why my initial method of solving this problem did yield an incorrect answer.

I began by creating an equation and drawing a right triangle.

x is the horizontal part of the triangle, y is the vertical part of the triangle, h is the hypotenuse.

I wrote the following equation:

(1/3)x=y

multiple both sides by (1/x)

(y/x)=1/3

tan(theta)=1/3

acrtang(1/3)=theta

My solution is wrong, but I don't know why.

Can someone please explain why my reasoning is not correct and doesn't yield the right answer?

 

[Doc Al]

Can someone please explain why my reasoning is not correct and doesn't yield the right answer?

Your solution assumes that the projectile travels in a straight line. It doesn't.

Instead, express the max height and range in terms of the projection angle.

 

[DEvens]

Also, don't forget the projectile first rises, then falls. There is an important factor of 2.

 

[kuruman]

[*** Quote removed at the request of @Klystron who deleted the post being quoted ***]

How do you figure? As @Doc Al observed,

Your solution assumes that the projectile travels in a straight line. It doesn't.

 

[DEvens]

My solution is wrong, but I don't know why.

How do you know it's wrong? What is it you are comparing to?

 

[MrDickinson]

How do you know it's wrong? What is it you are comparing to?

Dear Sir or Madam:

I did all of my work, then I checked the answer in the back of the book (this does not give a solution, just a plain answer).

With the wrong answer in mind, I really wanted to understandand why my reasoning was incorrect so that I understand the problem with greater depth.

The answers here have been very helpful.

Thanks

 

[MrDickinson]

How do you figure? As @Doc Al observed,

Thank you.

 

[MrDickinson]

Your solution assumes that the projectile travels in a straight line. It doesn't.

Instead, express the max height and range in terms of the projection angle.

Thank you. That is very helpful information.

Is it fair to say that the angle of projection is, with respect to the position of the object, is constantly changing?

Thanks

 

[Doc Al]

Is it fair to say that the angle of projection is, with respect to the position of the object, is constantly changing?

The angle of projection is just the angle that the projectile is launched -- the angle its initial velocity vector makes with the horizontal. The object's velocity changes as it rises and falls and the angle it makes changes as well.

 

[kuruman]

Here is a picture from the internet that shows the projection angle $\theta$. It is constant. What is changing continuously is the angle that the velocity vector forms with respect to the horizontal. At maximum height that angle is zero because the projectile is traveling parallel to the ground.