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(a) Using t- O as the time the stone leaves the thrower's hand at position A, determine the time at which the stone reaches its maximum height. SOLUTION You most likely have experience with dropping objects or throwing them upward and watching them fall, so this problem should describe a familiar experience. To simulate this situation, toss a small object upward and notice the time interval required for it to fall to the floor. Now imagine throwing that object upward from the roof of a building. Because the stone is in free fall, it is modeled as a particle under constant Select- V due to gravity. Recognize that the initial velocity is positive because the stone is launched upward. The velocity will change sign after the stone reaches its highest point, but the acceleration of the stone will always be -Select V so that it will always have aSelect-V value. Choose an initial point just after the stone leaves the person's hand and a final point at the top of its flight (Use the following as necessary: v anda Use the following equation to calculate the time at which the stone reaches its maximum height: Substitute numerical values. (Enter your answer in s.)

(a) Using t- O as the time the stone leaves the thrower's hand at position A, determine the time at which the stone reaches its maximum height. SOLUTION You most likely have experience with dropping objects or throwing them upward and watching them fall, so this problem should describe a familiar experience. To simulate this situation, toss a small object upward and notice the time interval required for it to fall to the floor. Now imagine throwing that object upward from the roof of a building. Because the stone is in free fall, it is modeled as a particle under constant Select- V due to gravity. Recognize that the initial velocity is positive because the stone is launched upward. The velocity will change sign after the stone reaches its highest point, but the acceleration of the stone will always be -Select V so that it will always have aSelect-V value. Choose an initial point just after the stone leaves the person's hand and a final point at the top of its flight (Use the following as necessary: v anda Use the following equation to calculate the time at which the stone reaches its maximum height: Substitute numerical values. (Enter your answer in s.)

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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(b) Find the maximum height of the stone. SOLUTION As in part (a), choose the initial and final points at the beginning and the end of the upward flight. Set y, = 0 and substitute the time from part (a) into the following equation to find the maximum height (in m): Y max = YB = YA xa² + (c) Determine the velocity of the stone when it returns to the height from which it was thrown. SOLUTION Choose the initial point where the stone is launched and the final point when it passes this position coming down. Substitute known values into the following equation. (Enter your answer in m/s.) 2a (Yc - Y) m/s When taking the square root, we could choose either a positive or a negative root. We choose the --Select-- root because we know that the stone is moving downward at point C. The velocity of the stone when it arrives back at its original height is equal in magnitude to its initial velocity but is opposite in direction. (d) Find the velocity and position of the stone at t = 5.00 s. SOLUTION Choose the initial point just after the throw and the final point 5.00 s later. Calculate the velocity (in m/s) at D from the following equation: VYD = VyA + ayt = 20.2 m/s + (-9.80 m/s2)(5.00 s) m/s Use the equation to find the position of the stone (in m) at t, = 5.00 s: YD = YA+ YA + = 0 + (20.2 m/s)(5.00 s) (-9.80 m/s?)(5.00 s)? m The choice of the time defined as t = 0 is arbitrary and up to you to select as the problem solver. As an example of this arbitrariness, choose t = 0 as the time at which the stone is at the highest point in its motion. Then solve parts (c) and (d) again using this new initial instant and note that your answers are the same as those above. EXERCISE Imagine an astronaut on an extrasolar planet, standing on a sheer cliff 50.0 m high. She is so happy to be on a different planet, she throws rock straight upward with an initial velocity of 15 m/s. If the astronaut were instead on Earth, and threw a ball in the same way while standing on a 50.0 m high cliff, what would be the time difference (in s) for the rock to hit the ground below the cliff in each case? Note that the gravitational acceleration on this planet is 4.4 m/s. Hint The time would be --Select--- on the extrasolar planet.
Transcribed Image Text:(b) Find the maximum height of the stone. SOLUTION As in part (a), choose the initial and final points at the beginning and the end of the upward flight. Set y, = 0 and substitute the time from part (a) into the following equation to find the maximum height (in m): Y max = YB = YA xa² + (c) Determine the velocity of the stone when it returns to the height from which it was thrown. SOLUTION Choose the initial point where the stone is launched and the final point when it passes this position coming down. Substitute known values into the following equation. (Enter your answer in m/s.) 2a (Yc - Y) m/s When taking the square root, we could choose either a positive or a negative root. We choose the --Select-- root because we know that the stone is moving downward at point C. The velocity of the stone when it arrives back at its original height is equal in magnitude to its initial velocity but is opposite in direction. (d) Find the velocity and position of the stone at t = 5.00 s. SOLUTION Choose the initial point just after the throw and the final point 5.00 s later. Calculate the velocity (in m/s) at D from the following equation: VYD = VyA + ayt = 20.2 m/s + (-9.80 m/s2)(5.00 s) m/s Use the equation to find the position of the stone (in m) at t, = 5.00 s: YD = YA+ YA + = 0 + (20.2 m/s)(5.00 s) (-9.80 m/s?)(5.00 s)? m The choice of the time defined as t = 0 is arbitrary and up to you to select as the problem solver. As an example of this arbitrariness, choose t = 0 as the time at which the stone is at the highest point in its motion. Then solve parts (c) and (d) again using this new initial instant and note that your answers are the same as those above. EXERCISE Imagine an astronaut on an extrasolar planet, standing on a sheer cliff 50.0 m high. She is so happy to be on a different planet, she throws rock straight upward with an initial velocity of 15 m/s. If the astronaut were instead on Earth, and threw a ball in the same way while standing on a 50.0 m high cliff, what would be the time difference (in s) for the rock to hit the ground below the cliff in each case? Note that the gravitational acceleration on this planet is 4.4 m/s. Hint The time would be --Select--- on the extrasolar planet.
A stone thrown from the top of a building is given an initial velocity of 20.2 m/s straight upward. The stone is launched 50.0 m above the ground, and the stone just misses the edge of the roof on its way down as shown in the figure below. B te = ? a,B = -9.80 m/s? LA = 0 YA = 0 a, A = -9.80 m/s2 c le = 2t8 Vc = 0 a,c = -9.80 m/s? to = 5.00 s 4D = -9.80 m/s2 YE = -H aE = -9.80 m/s? Position, velocity, and acceleration values at various times for a falling stone thrown initially upward with a velocity Vvi = 20.2 m/s. Many of the quantities in the labels for points in the motion of the stone are calculated in the example. Can you verify the other values that are not? (a) Using t, = 0 as the time the stone leaves the thrower's hand at position A, determine the time at which the stone reaches its maximum height. SOLUTION You most likely have experience with dropping objects or throwing them upward and watching them fall, so this problem should describe a familiar experience. To simulate this situation, toss a small object upward and notice the time interval required for it to fall to the floor. Now imagine throwing that object upward from the roof of a building. Because the stone is in free fall, it is modeled as a particle under constant --Select--- V due to gravity. Recognize that the initial velocity is positive because the stone is launched upward. The velocity will change sign after the stone reaches its highest point, but the acceleration of the stone will always be --Select--- top of its flight. so that it will always have a ---Select- V value. Choose an initial point just after the stone leaves the person's hand and a final point at the (Use the following as necessary: v yr Vyp x' and Use the following equation to calculate the time at which the stone reaches its maximum height: Vyr = Vyi + at→t = Substitute numerical values. (Enter your answer in s.) t = t8 =
Transcribed Image Text:A stone thrown from the top of a building is given an initial velocity of 20.2 m/s straight upward. The stone is launched 50.0 m above the ground, and the stone just misses the edge of the roof on its way down as shown in the figure below. B te = ? a,B = -9.80 m/s? LA = 0 YA = 0 a, A = -9.80 m/s2 c le = 2t8 Vc = 0 a,c = -9.80 m/s? to = 5.00 s 4D = -9.80 m/s2 YE = -H aE = -9.80 m/s? Position, velocity, and acceleration values at various times for a falling stone thrown initially upward with a velocity Vvi = 20.2 m/s. Many of the quantities in the labels for points in the motion of the stone are calculated in the example. Can you verify the other values that are not? (a) Using t, = 0 as the time the stone leaves the thrower's hand at position A, determine the time at which the stone reaches its maximum height. SOLUTION You most likely have experience with dropping objects or throwing them upward and watching them fall, so this problem should describe a familiar experience. To simulate this situation, toss a small object upward and notice the time interval required for it to fall to the floor. Now imagine throwing that object upward from the roof of a building. Because the stone is in free fall, it is modeled as a particle under constant --Select--- V due to gravity. Recognize that the initial velocity is positive because the stone is launched upward. The velocity will change sign after the stone reaches its highest point, but the acceleration of the stone will always be --Select--- top of its flight. so that it will always have a ---Select- V value. Choose an initial point just after the stone leaves the person's hand and a final point at the (Use the following as necessary: v yr Vyp x' and Use the following equation to calculate the time at which the stone reaches its maximum height: Vyr = Vyi + at→t = Substitute numerical values. (Enter your answer in s.) t = t8 =
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