Deriving an equation to find theta: Projectile motion

[Mac5214]

1. the question is this:

a projectile's speed at launch is four times its speed at the top of the trajectory. What is the launch angle?

I am given no numerical values, but from my understanding gravity is a constant 9.8, and at the maximum height velocity should be equal to 0. 2.the equation to find the maximum height is Hmax=Vi^2sin^2(theta)/2G and the equation for3. since the whole problem is diriving a equation i am lost as to where to start. other than knowing that the angle for maximum range is 45 degrees and that i can use the inverse of tan to find a angle of projection, but i am unsure how to find an angle that gives four times the initial speed. sorry if the question is sort of vague, it is all i was given. and thank you for any help you may provide.

 

[lekh2003]

You really need to list down all the kinematic equations. All of them. Then assess which ones are useful. Take those aside.

Then, think about what variables are changing and which are relevant.

In this case, you need to find how much time it takes for the projectile to reach the maximum height. It's good that you know how to derive that. But using that equation, plug that into the equation for speed at that point in time. I'm sure you should know what that is (even though it might take a little work).

Once you do that, you should be able to solve the rest.

Go through it slowly, we won't give you the answer, but we will make sure you're right along the way.

Have a go.

 

[CWatters]

Welcome to the forum Matthiasc. There is a template for posts in the homework section.

Please list any other Relevant Equations you know and your attempt at a solution so we can see where you are getting stuck.

I am given no numerical values, but from my understanding gravity is a constant 9.8, and at the maximum height vertical velocity should be equal to 0.

In case you didn't spot it the question 1. asks about speed not velocity.

For question 1 you don't actually need the equations of motion, just trig. Perhaps make a sketch and mark the speeds and angle on it.

 

[Mac5214]

Okay so I may have got a answer, but I didn’t use any kinematic equations so I’m not sure if it is correct or not, I realized that the horizontal component of velocity is constant, so when the projectile is at the maximum height, the vertical component of velocity will be equal to 0 therefore you could pick out any number for the x component of velocity and multiply it by 4 for the y component , and then do the inverse cos of the x component divided by the y component to find theta? So for instance I did cos-1(1/4) and got a angle of 75.5. Is my logic wrong in any place?

 

[haruspex]

you need to find how much time it takes for the projectile to reach the maximum height

No, it is irrelevant. Indeed, dimensional analysis shows the value of g must be irrelevant, so times and distances are not going to matter either.

 

[haruspex]

cos^-1(1/4)

Well done.

 

[tnich]

Okay so I may have got a answer, but I didn’t use any kinematic equations so I’m not sure if it is correct or not, I realized that the horizontal component of velocity is constant, so when the projectile is at the maximum height, the vertical component of velocity will be equal to 0 therefore you could pick out any number for the x component of velocity and multiply it by 4 for the y component , and then do the inverse cos of the x component divided by the y component to find theta? So for instance I did cos-1(1/4) and got a angle of 75.5. Is my logic wrong in any place?

Your numerical answer is correct. To get $cos(\theta)$, you need to divide the horizontal component by the hypotenuse, not the vertical component. Do you see why this is what you really did?

 

[Mac5214]

Your numerical answer is correct. To get $cos(\theta)$, you need to divide the horizontal component by the hypotenuse, not the vertical component. Do you see why this is what you really did?

oh okay yes i do, its because cosine is the adjacent over the hypotenuse which is the total magnitude (4).
Thank you for helping me realize that.