(b) Find the maximum height of the stone.SOLUTIONAs 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 = YAxa² +(c) Determine the velocity of the stone when it returns to the height from which it was thrown.SOLUTIONChoose 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/sWhen 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.SOLUTIONChoose 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/sUse 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)?mThe 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.EXERCISEImagine 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 throwsrock 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 hitthe ground below the cliff in each case? Note that the gravitational acceleration on this planet is 4.4 m/s.HintThe 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.Bte = ?a,B = -9.80 m/s?LA = 0YA = 0a, A = -9.80 m/s2c le = 2t8Vc = 0a,c = -9.80 m/s?to = 5.00 s4D = -9.80 m/s2YE = -HaE = -9.80 m/s?Position, velocity, and acceleration values at various times for afalling stone thrown initially upward with a velocityVvi = 20.2 m/s. Many of the quantities in the labels for points inthe motion of the stone are calculated in the example. Can youverify 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.SOLUTIONYou 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 thestone 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: vyr Vyp x'andUse 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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