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An object is dropped from a tower, 1936 ft above the ground. The object's height above the ground t seconds after it is dropped is

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An object is dropped from a tower, 1936 ft above the ground. The object's height above the ground t seconds after it is dropped is s(t) = 1936-16t". Determine the velocity and acceleration of the object the moment it reaches the ground. The velocity of the object the moment it reaches the ground is | f/s. (Simplify your answer.) The acceleration of the object the moment it reaches the ground is fu/s2. (Simplify your answer.)350t - 501- A small probe is launched vertically from the ground. After it reaches its high point, a parachute deploys and the probe descends to Earth. The height of the probe above ground is s(1) = -, for 0 sts7. 1 + 1 a. Graph the height function and describe the motion of the probe. b. Find the velocity of the probe. c. Graph the velocity function and determine the approximate time at which the velocity is a maximum. a. Choose the correct graph of the height function below. O A. O B. O C. OD. Q [0,7,1] by [0, 180,5] [0,7,1] by [0,360, 10] [0,7,1] by [0, 180,5] [0,7,1] by [0,180,5] Describe the motion of the probe. Choose the correct answer below. O A. The probe launches upward at a decreasing rate over the first half of its trip. Then it reaches a peak and falls at a decreasing rate until it reaches the ground at 7 seconds. O B. The probe launches upward moving fastest at the beginning and slows over the first second. It peaks and begins to fall downward slowing as it gets closer to the ground until reaching the ground at 7 seconds. O C. The probe launches at a fast speed and slows at a decreasing rate over the first half of its trip. It begins to fall at an increasing rate until it reaches that ground at 7 seconds O D. The probe speeds up quickly over the first second, then begins to slow down over the first half of its trip. Then, the probe begins to fall at an increasing rate until the last second where it begins to slow down until it reaches the ground at 7 seconds. b. v(t) = c. Choose the correct graph of the velocity function below. O A. O B. O C. OD. Q [0,7,1] by [ - 180, 180, 10] [0,7,1] by [ - 180,370, 10] [0,7,1] by [ - 50,50,5] 10,7,1] by [ - 370, 180,10] The velocity is at a maximum at approximately t= ]- (Type a whole number.)The accompanying graph shows the position of a bicyclist after t hours. Assume s is positive when the bicyclist is east of his starting point. a. Determine the average velocity of the bicyclist over the first hour of the trip. Q O A. - 60 mph O B. 60 mph O C. 10 mph O D. - 10 mph Position (miles after t hours) b. Find the average velocity of the bicyclist over the time interval [1,2.5] O A. - 32 mph O B. -6.7 mph Time (hours) O C. 6.7 mph Using the graph, answer parts a through d. O D. 32 mph Is the average velocity a good estimate of the instantaneous velocity at t = 2 hrs? O No Yes c. Find the average velocity of the bicyclist over the time interval [2,3.5]. O A. - 12 mph O B. 12 mph O C. -4 mph O D. 4 mph Estimate the velocity of the bicyclist at t= 0.5 hours and determine the direction in which the bicyclist is moving. O A. 10 mph to the east O B. 2 mph to the west O C. 60 mph to the east O D. 10 mph to the west d. Describe the movement of the bicyclist during the 4 hour trip. O A. The bicyclist rides up a hill the entire time. O B. The bicyclist rides to the east, takes a break after two hours, and then rides to the west. O C. The bicyclist takes a curvy path, riding both to the east and to the west O D. The bicyclist rides to the east, takes a break after two hours, and then rides to the east again

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