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.

Question

(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!

Answer 1

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