Projectile Motion Analysis: Time in Air Calculation

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Homework Help Overview

The discussion revolves around projectile motion, specifically analyzing the time a projectile remains in the air when fired horizontally from a height of 45.0 m with an initial speed of 250 m/s. Participants explore the relevant equations and concepts related to free fall and horizontal motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss various kinematic equations and their applicability to the problem. There is an exploration of the relationship between horizontal and vertical motion, with some questioning the relevance of horizontal velocity in determining time of flight. Others suggest using free fall equations to find the time based on vertical distance.

Discussion Status

The discussion is active, with participants providing insights and corrections to each other's reasoning. Some have offered guidance on using specific equations, while others are clarifying the implications of horizontal motion on the time of flight. Multiple interpretations of the problem are being explored, particularly regarding the role of horizontal velocity.

Contextual Notes

Participants note that the problem involves assumptions about neglecting air resistance and the nature of horizontal motion, which may affect the calculations and interpretations of the projectile's behavior.

  • #31
The magnitude of the initial velocity.
 
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  • #32
Blox_Nitrates said:
The magnitude of the initial velocity.
No, that's not how it works. Consider horizontal and vertical components of velocity separately. The vertical component is subject to acceleration by gravity; the horizontal one is not. So, ignoring air resistance, the horizontal component of velocity does not change. If it is V at max height, what is it at launch?
 
  • #33
The horizontal component would also be V at the initial launch if ignoring air resistance because velocity wouldn't change.
 
  • #34
Since we are ignoring air resistance than the initial speed v would be the same as initial speed v at max height because the horizontal component's velocity doesn't change. The speed at launch would also be speed V.
 
  • #35
Blox_Nitrates said:
Since we are ignoring air resistance than the initial speed v would be the same as initial speed v at max height because the horizontal component's velocity doesn't change. The speed at launch would also be speed V.
The horizontal component of the velocity does not change, but the projectile's speed (which combines horizontal and vertical components) does.
Suppose the launch speed is u and launch angle is θ above horizontal. What is the horizontal component at launch?
 
  • #36
haruspex said:
The horizontal component of the velocity does not change, but the projectile's speed (which combines horizontal and vertical components) does.
Suppose the launch speed is u and launch angle is θ above horizontal. What is the horizontal component at launch?

Voy = usinθ?
 
  • #37
Blox_Nitrates said:
Voy = usinθ?
That is correct for the vertical velocity at launch, but I asked for the horizontal.
 
  • #38
haruspex said:
That is correct for the vertical velocity at launch, but I asked for the horizontal.

Sorry,
Vox = ucosθ
 
  • #39
Right, so let's recap. The question was:
A projectile's launch speed is five times its speed at maximum height. Find launch angle θo
We have the variable v for speed at max height, we have established that this is equal to Vox = ucosθo. What does the question say about the relationship between u and v?
 

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