ATAR Notes: Forum
VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: shokstar on May 01, 2010, 08:05:57 pm
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Hi everyone, im a bit confused as to the right approach to the type of projectile motion that i posted,
For instance, a ball is thrown from a cliff xm above the ground at an angle of y (is an angle needed?) with a velocity of vms-1, what is the max height it reaches above the ground, how far from the base of the cliff does it land, and what is the time?
What else is needed? My book, Jac-shit, doesnt cover this type of situation, and my teacher hasnt really covered it. Any help would be appreciated.
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im a bit confused, wat is it that u are after?
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a good way is to split motion into horizontal and vertical components, then apply your normal kinematics formulas.
so eg, the horizontal component the acceleration would be 0.
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What about the first part of its journey? The part where it is shot, and the period before it crosses the level of the cliff again. Does that not mean it needs a unique approach? Or is it just the same as normal projectile motion?
I was under the impression that the first part made it necessary for a different approach.
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i'd say same type of approach
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Ok thanks mate.
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You need the angle to the get the x and y components of the velocity.
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yeah, thanks
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this
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Im having trouble finding the time. I have tried using v=u+at too...
Two stones are thrown simultaneously in line and horizontally from the top of a cliff 245m high with speeds 16m/s and 64m/s respectively.
a) How far apart will they strike the water
b) Find the speed of the first stone as it reaches the water. (is there another way other than using potential energy=kinetic energy?)
Cheers
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In the vertical direction the initial velocity is 0, the acceleration is -9.8, and the displacement is 245. So you can use the formula x=ut+(1\2)at^2 to find the time.
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thanks mate (Y)