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December 20, 2014

December 20, 2014

Posted by **Technoboi11** on Sunday, March 11, 2007 at 11:44pm.

(a) Using energy techniques, rather than techniques of Chapter 4, find the speed of the snowball as it reaches the ground below the cliff.

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ok so I used the equation k1+u1=k2=u2

plugging in numbers i did 1/2(1.50kg)(14m/s)^2+(1.50kg)(9.8m/s^2)(13.5m)=1/2(1.50kg)(vf)^2+(1.50kg)(9.8m/s)(0).

The answer is 21 m/s.

Well for part b the question says "What is that speed if, instead, the launch angle is 41.0° below the horizontal". How do I change my equation if the angle changing to a negative is not involved in the equation used???

The downward angle changes the initial vertical velocity to downward, or a sign change. Since it is squared, it makes no difference in the final velocity.

Ah well that was dumb for sure on my part. thanks for the help!

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