A 1.50 kg snowball is fired from a cliff 13.5 m high with an initial velocity of 14.0 m/s, directed 41.0° above the horizontal.
(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!
a 65-kg ice skater moving to the right with a velocity of 2.50m/s throws a .150kg snowball to the right with a velocity of 32.0m/s relative to the ground a) what is the velocity of the ice skater after throwing the snowball? b) a
asked by Sarah on April 26, 2012
You're welcome! Don't worry, we all make mistakes sometimes. I'm glad I could help clarify things for you. If you have any more questions, feel free to ask!
No problem! It's not a dumb question at all. When the launch angle changes to a negative angle, it simply means that the direction of the initial velocity is now pointing downwards instead of upwards. In this case, the downward angle doesn't affect the equation you used because the equation you used only accounts for the initial and final kinetic energies and gravitational potential energy.
Since the downward launch angle only affects the direction of the initial velocity, the equation remains the same:
1/2(1.50 kg)(vi)^2 + (1.50 kg)(9.8 m/s^2)(13.5 m) = 1/2(1.50 kg)(vf)^2 + (1.50 kg)(9.8 m/s^2)(0)
In both cases, whether the launch angle is positive or negative, the equation includes the gravitational potential energy (mgh) at the starting point and sets it equal to the final kinetic energy (1/2mvf^2) at the landing point.
I hope that clarifies your question! Let me know if you have any more doubts.