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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- Thread Starter
- 16-10-2016 16:50
- Official Rep
- 18-10-2016 18:03
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Just quoting in Danny Dorito so she can move the thread if needed
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- Political Ambassador
- 18-10-2016 18:05
oh misread the question
- 18-10-2016 18:07
- 18-10-2016 18:12
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.Last edited by Callicious; 18-10-2016 at 18:14.
- 18-10-2016 19:20
Vertical velocity at top of flight=0
Posted from TSR Mobile
- 18-10-2016 19:22
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.