Projectile Motion Analysis: Time in Air Calculation

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A projectile fired horizontally from a height of 45.0 meters with a speed of 250 m/s remains in the air for approximately 3.03 seconds, as determined by analyzing the free fall motion. The horizontal speed is extraneous to the time calculation, which is solely dependent on the vertical distance and gravitational acceleration. To find the horizontal distance traveled during this time, the formula distance = speed × time yields approximately 758 meters. Additionally, discussions about projectile motion clarify that at maximum height, the projectile's vertical velocity is zero while the horizontal component remains unchanged. The launch speed is five times the speed at maximum height, leading to further analysis of the launch angle and velocity components.
  • #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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