What is the Initial Projection Angle of a Projectile with Varied Speed Heights?

[pereus]

Homework Statement

The speed of a projectile when it reaches its maximum height is 1/4 the speed when the projectile is at 1/4 its maximum height.
What is the initial projection angle?
Answer in units of ◦

Homework Equations

v² = v₀² + 2a(delta s)
4V_x = v_1/4

The Attempt at a Solution

I've tried substituting v_1/4 for 4v_x and some other things, but I lost those papers.

 

[PeterO]

Homework Statement

The speed of a projectile when it reaches its maximum height is 1/4 the speed when the projectile is at 1/4 its maximum height.
What is the initial projection angle?
Answer in units of ◦

Homework Equations

v² = v₀² + 2a(delta s)
4V_x = v_1/4

The Attempt at a Solution

I've tried substituting v_1/4 for 4v_x and some other things, but I lost those papers.

What is the significance of the speed at maximum height? What is the velocity at maximum height?

 

[pereus]

I know that the velocity at the top is V_x. That is how I was able to make 4v_x = to 4v_1/4

 

[PeterO]

I know that the velocity at the top is V_x. That is how I was able to make 4v_x = to 4v_1/4

What do you mean by V_x ? What is significant at the top?

 

[pereus]

At the top the projectile is moving with the velocity V_x which is equal to v₀cos(theta). There is no velocity in the y direction

 

[PeterO]

At the top the projectile is moving with the velocity V_x which is equal to v₀cos(theta). There is no velocity in the y direction

OK.

You can actually analyse the second half of the projectile flight, as it mirrors the first [going down and going up work the same numerically] however.

At 1/4 the max height, there will be a vertical component to the velocity, and application of Pythagorus will enable you to find that, since the result is 4V_x.
You will get the vertical component in terms of V_x, but you are only using it to find an angle.
Once you have the vertical vel 1/4 way up - or if you like, 3/4 way down, you can find it at the beginning / end and again use trig to find the angle.

You could actually be using Potential Energy / Kinetic Energy / Total Energy to be working this out.