Projectile Motion Analysis: Time in Air Calculation

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SUMMARY

The discussion focuses on calculating the time a projectile remains in the air when fired horizontally from a height of 45.0 meters with an initial speed of 250 m/s. The correct approach involves recognizing that the horizontal velocity is extraneous to the vertical motion, which is governed solely by gravity. The time in the air is calculated using the equation for free fall, yielding a result of 3.03 seconds. Additionally, the horizontal distance traveled during this time is calculated to be approximately 758 meters.

PREREQUISITES
  • Understanding of kinematic equations, specifically for free fall.
  • Familiarity with projectile motion concepts.
  • Knowledge of basic calculus principles as they apply to physics.
  • Ability to differentiate between horizontal and vertical components of motion.
NEXT STEPS
  • Study the derivation and application of the kinematic equation y = Vo*t + 0.5*a*t^2.
  • Learn about the effects of air resistance on projectile motion.
  • Explore the relationship between launch angle and range in projectile motion.
  • Investigate the concept of maximum height and its calculation in projectile motion.
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the principles of projectile motion and free fall calculations.

  • #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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