5. DETAILS M Grader Information: • This question provides no feedback on your answers. . You can change your answer as m

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5 Details M Grader Information This Question Provides No Feedback On Your Answers You Can Change Your Answer As M 1 (43.85 KiB) Viewed 36 times

5. DETAILS M Grader Information: • This question provides no feedback on your answers. . You can change your answer as many times as you need. . Only your final answer is graded. Be sure to submit your answer. An object is launched from a platform 400 feet (ft) above ground level. The trajectory of this object is shown in the graph below. v (ft) x (ft) Once the object has been launched, the position of the object can be described using the position vector function: (t) = (110t, -15.12 + 50t + 400) ft Use the drop-down menus below to indicate which of the following are true or false statements. Select-- To find when the object reaches its maximum height solve -16.12 + 50+ + 400 = 0. The acceleration of the object is (0,32.2) ft/s2. --Select-- Emax ---Select- The total distance this object traveled is (1101)2 + (-16.1? + 50t + 400)2dt. --Select- --Select- -- Select --Select- The initial velocity of the object is (110, 50) ft/s. The angle in which the object was launched into the air is tan-+(60/110). To find when the object hits the ground solve -32.2t + 50 = 0. The initial speed of the object is 1102 + 602 ft/s. When the object reaches its maximum height, the vertical component of its velocity is zero. --Select- --Select-