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Answer :
Final answer:
The speed of a 55kg lead ball after falling 39.1m from rest is approximately 27.69 m/s, obtained using the kinematic equation for an object in free fall.
Explanation:
To find the speed of the ball after falling 39.1 meters from rest, we can use the formula derived from the basic equations of kinematics for an object in free fall:
v2 = u2 + 2gh
Where v is the final velocity, u is the initial velocity (which is 0 m/s in this case), g is the acceleration due to gravity (9.81 m/s2 on Earth), and h is the height (39.1 m).
Plugging the values into the equation:
v2 = 02 + 2(9.81 m/s2)(39.1 m)
v2 = 766.962 m2/s2
v = √(766.962 m2/s2)
v = 27.69 m/s (rounded to two decimal places)
Therefore, after falling 39.1 meters, the speed of the lead ball would be approximately 27.69 m/s.
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