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Find the time required for the stone to reach the nd.

Find the time required for the stone to reach the nd.

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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口 Q Example 2.9 Not a Bad Throw for a Rookie! Goal Apply the kinematic equations to a freely falling object with a nonzero initial velocity. Problem A stone is thrown from the top of a building with an initial velocity of 24.9 m/s straight upward, at an initial height of 56.7 m above the ground. The stone just misses the edge of the roof on its way down, as shown in Figure 2.20. (a) Determine the time needed for the stone to reach its maximum height. (b) Determine the maximum height. (c) Determine the time needed for the stone to return to the height from which it was thrown, and the velocity of the stone at that instant. (d) Determine the time needed for the stone to reach the ground. (e) Determine the velocity and position of the stone at t 5.55 s. Strategy The diagram in Figure 2.20 establishes a coordinate system with Yo 0 at the level of release from the thrower's hand, with y positive upward. Write the velocity and position kinematic equations for the stone and substitute the given information. All the answers come from these two equations by using simple algebra or just substituting the time. For part (a), for example, the stone comes to rest for an instant at maximum height, so set v = 0 at this point and solve for time. Then substitute the time into the displacement equation, obtaining the maximum height. Figure 2.20 A freely falling object is thrown upward from a building.
Transcribed Image Text:口 Q Example 2.9 Not a Bad Throw for a Rookie! Goal Apply the kinematic equations to a freely falling object with a nonzero initial velocity. Problem A stone is thrown from the top of a building with an initial velocity of 24.9 m/s straight upward, at an initial height of 56.7 m above the ground. The stone just misses the edge of the roof on its way down, as shown in Figure 2.20. (a) Determine the time needed for the stone to reach its maximum height. (b) Determine the maximum height. (c) Determine the time needed for the stone to return to the height from which it was thrown, and the velocity of the stone at that instant. (d) Determine the time needed for the stone to reach the ground. (e) Determine the velocity and position of the stone at t 5.55 s. Strategy The diagram in Figure 2.20 establishes a coordinate system with Yo 0 at the level of release from the thrower's hand, with y positive upward. Write the velocity and position kinematic equations for the stone and substitute the given information. All the answers come from these two equations by using simple algebra or just substituting the time. For part (a), for example, the stone comes to rest for an instant at maximum height, so set v = 0 at this point and solve for time. Then substitute the time into the displacement equation, obtaining the maximum height. Figure 2.20 A freely falling object is thrown upward from a building.
O < O Q Q D # n 5 e (a) Find the time when the stone reaches its maximum height. Write the velocity and position kinematic equations. v = at + vo Ay y Yo = Vot + at Substitute a = -9.80 m/s, Vo = 24.9 m/s, and yo = 0 into the preceding two equations. v = (-9.80 m/s²)t + 24.9 m/s (1) y (24.9 m/s)t - (4.90 m/s2)t2 (2) Substitute v = 0, the velocity at maximum height, into Equation (1) and solve for time. 0 - (-9.80 m/s2)e + 20.0 m/s -24.9m t 3= -9.80m/s 2.548 (b) Determine the stone's maximum height. Substitute the time found in part (a) into Equation (2). Ymax = (24.9 m/s)(t s) - (4.90 m/s?)(t s)? = 31.6 (c) Find the time the stone takes to return to its initial position, and find the velocity of the stone at that time. Set y = 0 in Equation (2) and solve for t. 0 (24.9 m/s)t-(4.90 m/s)t = t(24.9 m/s - 4.90 m/st) t = 5.08 Substitute the time into Equation (1) to get the velocity. v = 24.9 m/s + (-9.80 m/s)(e) -24.9 V m/s V (d) Find the time required for the stone to reach the ground. In Equation (2), set y = -56.7 m. -56.7 m = (-24.9 m/s)t- (4.90 m/s)e Apply the quadratic formula and take the positive root. t=6.7866
Transcribed Image Text:O < O Q Q D # n 5 e (a) Find the time when the stone reaches its maximum height. Write the velocity and position kinematic equations. v = at + vo Ay y Yo = Vot + at Substitute a = -9.80 m/s, Vo = 24.9 m/s, and yo = 0 into the preceding two equations. v = (-9.80 m/s²)t + 24.9 m/s (1) y (24.9 m/s)t - (4.90 m/s2)t2 (2) Substitute v = 0, the velocity at maximum height, into Equation (1) and solve for time. 0 - (-9.80 m/s2)e + 20.0 m/s -24.9m t 3= -9.80m/s 2.548 (b) Determine the stone's maximum height. Substitute the time found in part (a) into Equation (2). Ymax = (24.9 m/s)(t s) - (4.90 m/s?)(t s)? = 31.6 (c) Find the time the stone takes to return to its initial position, and find the velocity of the stone at that time. Set y = 0 in Equation (2) and solve for t. 0 (24.9 m/s)t-(4.90 m/s)t = t(24.9 m/s - 4.90 m/st) t = 5.08 Substitute the time into Equation (1) to get the velocity. v = 24.9 m/s + (-9.80 m/s)(e) -24.9 V m/s V (d) Find the time required for the stone to reach the ground. In Equation (2), set y = -56.7 m. -56.7 m = (-24.9 m/s)t- (4.90 m/s)e Apply the quadratic formula and take the positive root. t=6.7866
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