Projectile Motion Analysis: Time in Air Calculation

In summary: It doesn't mean that the angle is chosen to achieve maximum height. The angle is whatever it is. mfb is asking what the direction of motion of the projectile will be when the projectile is at its maximum height on the trajectory. Sketching it might help.
  • #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?
 
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  • #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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