2. A 0.18kg baseball is thrown straight up with a speed of 35m/s. By the time it
ID: 2024278 • Letter: 2
Question
2. A 0.18kg baseball is thrown straight up with a speed of 35m/s. By the time it reaches the height of 38m the speed of the ball has been reduced to 12m/s.
A. Determine the average force of air resistance during this trip.
B. How many joules of waste heat are produced as the ball comes to rest at its peak at 43.1m while slowing down from 12m/s?
C. If the ball falls to the earth from the peak height with the same average force of air resistance as you found in part A, what will the impact speed be?
D. Is the impact speed the same as the speed with which the ball was thrown? Why or why not?
Answer: A - 0.796 N force of air resistance, B - 3.96 J of waste heat
My issue is how am I supposed to solve the Vf if I don't know the total distance...
Explanation / Answer
Seems to me that they give you the starting height of 43.1 m in part B. To solve part C, just look at the acceleration.
The ball has a weight of (0.18 kg)(9.81 m/s2) = 1.7658 N. That force is pulling it down, and your average force of air resistance 0.796 N is holding it back (I'll take your word for it that that is the correct number -- and you know in real life air resistance is proportional to velocity, not constant) so the net force acting downward is 0.9698 N.
a = F/m so the downward acceleration iis 5.388 m/s2. t = (2d/a)1/2 = (2 * 43.1m)/(5.388 m/s2)1/2
= 4.00 s time to fall, so final velocity is at = 5.388 m/s2 * 4.00 s = 21.55 m/s.
I think that's right. And no, that isn't the same speed as it went up, because it lost energy to air resistance.
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