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A projectile is fired from a seige engine at 35ms at an angle of 60 degrees to the horizonal.
a)what is its vertical velocity at the top of its flight?
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Reply 2
oh misread the question
Reply 3
Original post by isthisgood
A projectile is fired from a seige engine at 35ms at an angle of 60 degrees to the horizonal.
a)what is its vertical velocity at the top of its flight?


Do you know how to resolve a vector into components?
Original post by isthisgood
A projectile is fired from a seige engine at 35ms at an angle of 60 degrees to the horizonal.
a)what is its vertical velocity at the top of its flight?


You'll need to resolve the components first; we can do this through pythagoras! Our best friend in the whole world.

Imagine you've got a triangle with one side against a wall, one side over the roof, then an angle from the roof down to our velocity which is a line from the corner to the middle of the room. Sort of like this. The *** shows the angle, the top is our horizontal component, and the side our vertical
________________
|-+---------*
|---+------*
|-----+--*
|--------+
|----------+

Our +++ line is our full velocity, or actual, whichever you want to call it, 35 ms. We can resolve it by 35cos(phi), where phi in this situation is *** which is 60 degrees (for horizontal), or 35sin(phi) for our vertical component.

So, 35 sin 60 is our vertical component, which should be 30.31 ms.
(edited 7 years ago)
Vertical velocity at top of flight=0


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Reply 6
At the top of its flight the projectile is instantaneously at rest (vertically), if you draw its path, the top of its path represents the point where it is momentarily changing direction from going upwards to falling downwards.

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