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Chapter 2: Problem 51

please assume the free-fall acceleration \(g=9.80 \mathrm{m} / \mathrm{s}^{2}\) unless a more precise value is given in the problem statement. Ignore air resistance. A 55 -kg lead ball is dropped from the leaning tower of Pisa. The tower is 55 m high. (a) How far does the ball fall in the first 3.0 s of flight? (b) What is the speed of the ball after it has traveled 2.5 m downward? (c) What is the speed of the ball 3.0 s after it is released? (d) If the ball is thrown vertically upward from the top of the tower with an initial speed of $4.80 \mathrm{m} / \mathrm{s}$, where will it be after 2.42 s?

Short Answer

Expert verified
Answer: The ball falls 44.1 meters in the first 3.0 seconds, its speed after falling 2.5 meters is 7.0 m/s, its speed 3.0 seconds after being released is 29.4 m/s, and its location after 2.42 seconds if thrown vertically upward with an initial speed of \(4.80 \mathrm{m} / \mathrm{s}\) would be approximately 53.81 meters above the ground.

Step by step solution

01

(a) Distance fallen in the first 3.0 s

Since the ball is dropped, its initial velocity \((u)\) is 0. We can use the equation \(s = ut + \frac{1}{2}at^2\) to find the distance fallen. Plugging in the given values: \(s = (0)(3) + \frac{1}{2}(9.80)(3)^2 = \frac{1}{2}(9.80)(9) = 44.1 \mathrm{m}\) The ball falls 44.1 meters in the first 3 seconds.
02

(b) Speed of the ball after falling 2.5 m

To find the speed of the ball after falling 2.5 m, we can use the equation \(v^2 = u^2 + 2as\). Since the ball is dropped, its initial velocity \((u)\) is 0. Plugging in the given values: \(v^2 = (0)^2 + 2(9.80)(2.5) = 49.0\) \(v = \sqrt{49.0} = 7.0 \mathrm{m/s}\) The speed of the ball after falling 2.5 meters is 7.0 m/s.
03

(c) Speed of the ball 3.0 s after release

To find the speed of the ball 3.0 s after release, we can use the equation \(v = u + at\). Since the ball is dropped, its initial velocity \((u)\) is 0. Plugging in the given values: \(v = 0 + (9.80)(3) = 29.4 \mathrm{m/s}\) The speed of the ball 3.0 seconds after release is 29.4 m/s.
04

(d) Location of the ball 2.42 s after being thrown upward

To find the location of the ball 2.42 s after being thrown upward with an initial speed of \(4.80 \mathrm{m} / \mathrm{s}\), we need to consider that the acceleration \((a)\) will be negative (-9.80 m/s²) since it is opposing gravity. We can use the equation \(s = ut + \frac{1}{2}at^2\) to find the displacement from the top of the tower. Plugging in the given values: \(s = (4.80)(2.42) + \frac{1}{2}(-9.80)(2.42)^2 \approx 1.1856 \,\mathrm{m}\) To find the position of the ball from the ground, subtract this value from the height of the tower: \(55 - 1.1856 \approx 53.8144 \,\mathrm{m}\) The ball will be approximately 53.81 meters above the ground after 2.42 seconds.

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