anonymous
  • anonymous
a stick is thrown from a cliff 27m high with an initial velocity of 18m/sat an angle of 37 above the horizontial a) use the law of conservation of energy to determine the speed of the stick just before it hits ground b) repeat (a) with an angle of 37 degrees below the horizontial
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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kunal
  • kunal
the final velocity will be in both cases be 25.63m/s
anonymous
  • anonymous
okay how'd u get there can u show me pls
anonymous
  • anonymous
btw my book get 29m/s

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anonymous
  • anonymous
@kunal Please do not just post answers... what are those good for? This is a learning community. Please take the time to walk people through how you get answers. For more information, please refer to the Code of Conduct: http://openstudy.com/code-of-conduct
kunal
  • kunal
aly33 ,,, firstly when we throw the stick it moves in a projectile motion.... and it is well known fact that the time taken by a ball while its journey upwards is same the time taken for downward journey.........since distance is same therefore the speed/velocity will be same....
anonymous
  • anonymous
First, let's determine the x and y-components of the initial velocity. \[v_x = v \cos(37) ~~{\rm and}~~ v_y = v \sin(\theta)\]If we are to ignore wind-resistance, we know that \(v_x\) will remain constant throughout the the entire flight since no forces act in the x-direction. We know that \(v_y\) will initially decrease as the projectile heads upwards. Once the projectile reaches the top of its trajectory (\(v_y=0\) here) it will accelerate towards the earth. Now, we need to determine how high the object is at the top of it's trajectory. We will use the launch point as the reference for this portion. From conservation of energy\[\Delta KE = \Delta PE\]which means that all the kinetic energy that the projectile possesses at launch will all be converted to potential energy when the projectile reaches the top of its trajectory. \[{1 \over 2} m (v_y)^2 = mg h \rightarrow {1 \over 2} v_y^2 = gh\]we know \(v_y\) and g, therefore we can solve for h. Now, remember this is the height that the projectile gained from its initial launch point. The launch point is 27m above the surface of the earth. Therefore, now we need to consider the potential energy of the projectile from the surface of the earth. Again, we can use conservation of energy \[{1 \over 2} v_{y,i}^2 = g(h+27)\]where h is the height we calculated previously and \(v_{y,i}\) is the velocity in the y-direction at the time of impact. We can solve for \(v_{y,i}\). Now, to determine the magnitude of velocity, we need to use Pythagorean's Theorem\[v_i = \sqrt{~v_x^2 + v_{y,i}^2}\] For part b, the vertical component of velocity can be found as\[{1 \over 2} v_{y,i}^2 = {1 \over 2} v_y^2 + gh \]where \(v_{y,i}\) is the impact velocity in the y-direction, \(v_y\) is the initial velocity, and h is the height of the cliff.
anonymous
  • anonymous
kunal. The stick is thrown from a cliff that is 27m high.
kunal
  • kunal
eashmore.. i know that the stick is trown from a cliff 27 m high so what it accounts for??
kunal
  • kunal
i m taking into consideration the basic equations of motion
anonymous
  • anonymous
It will gain energy as it returns to the earth by falling the additional 27m. Therefore, Energy at launch DOES NOT equal energy at impact. Therefore, the speeds will NOT be the same at launch and impact.
kunal
  • kunal
that i know........ but i am considering the point in sky at which the projectile will be at same at as the cliff.. i.e. 27m ........ the further journey will include a second projectile motion.
anonymous
  • anonymous
okay so i start by determining my x y components that makes sense, then heigh using the formula Ek=Ep this equals 1/2v^2=mgh therefore h=.5v^2/mg??? then i get a little confused
anonymous
  • anonymous
is h= to my y component? and then in the next equation i do y+27m?
kunal
  • kunal
there we have the velocity once again as 18m/s ....... there we take the vector. that is straight downwards. i.e. 18 sin37 and apply the equation \[v ^{2}-u ^{2}=2gh\] to get the final velocity.. in downward direction.. then find the vector sum of the horizontal and vertical vectors. that will be our final velocity. and it is 29.35m/s
anonymous
  • anonymous
Okay. It was unclear in your explanation where you reference was...... In the first equation. h is equal to the height gained by the projectile above the CLIFF surface. After it reaches its maximum height, we need to switch our reference point for potential energy to the ground. This is done by adding 27m to the height gained above the cliff.
anonymous
  • anonymous
Also, there should be no mass in the denominator in your equation for height.

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