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Kelsie Sawayn

Feb 20, 2025

A 0.025-kg bullet is fired vertically at 208 m/s into a 0.15-kg baseball that is initially at rest. The bullet?

A 0.025-kg bullet is fired vertically at 208 m/s into a 0.15-kg baseball that is initially at rest. The bullet lodges in the baseball and, after the collision, the baseball/bullet rise to a height of 37 m.What was the average force of air resistance while the baseball/bullet was rising? (Consider the positive direction to be upwards.)
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Madie Hilpert

Feb 20, 2025

Assume momentum is conserved in the collision and solve for the initial velocity of the bullet-ball…MV (bullet) = MV (total)(0.025)(208) = (0.025 + 0.15)VV = 29.71 m/sUse conservation of energy to find the height the bullet-ball would have risen without air resistance…½mv² = mghh = v² / 2gh = (29.71)² / (2)(9.81)h = 45.0 mThe difference in potential energy between the ideal height and the actual height is the amount of energy lost to work against air resistance….mgh2 – mgh1 = FΔh(0.175)(9.81)(45) – (0.175)(9.81)(37) = F(37)F = 0.371 N <=========================... Show More
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