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

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The discussion focuses on determining the initial projection angle of a projectile based on its speed at various heights. It is established that the speed at maximum height is one-fourth of the speed at one-fourth of the maximum height. The velocity at maximum height is identified as Vx, which equals v0cos(theta), with no vertical component at that point. Participants suggest using principles of energy conservation and trigonometry to analyze the projectile's motion and derive the angle. The analysis emphasizes that the projectile's behavior is symmetrical, allowing for calculations based on the vertical and horizontal components of velocity.
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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)
4Vx = v1/4

The Attempt at a Solution


I've tried substituting v1/4 for 4vx and some other things, but I lost those papers.
 
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pereus said:

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)
4Vx = v1/4

The Attempt at a Solution


I've tried substituting v1/4 for 4vx 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?
 
I know that the velocity at the top is Vx. That is how I was able to make 4vx = to 4v1/4
 
pereus said:
I know that the velocity at the top is Vx. That is how I was able to make 4vx = to 4v1/4

What do you mean by Vx ? What is significant at the top?
 
At the top the projectile is moving with the velocity Vx which is equal to v0cos(theta). There is no velocity in the y direction
 
pereus said:
At the top the projectile is moving with the velocity Vx 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 4Vx.
You will get the vertical component in terms of Vx, 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.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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