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Today in Precalculus Go over homework Notes: Simulating Projectile Motion Day 2 Homework

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Objects going straight up or down A baseball is hit straight up from a height of 3ft with an initial velocity of 40ft/sec. The parametric equations that model the height of the ball as a function of time t is: x = t y = -16t 2 + 40t + 3 What is the height of the ball after 0.5 seconds? y=-16(.5) 2 + 40(.5) + 3=19ft What is the maximum height of the ball and how long does it take to reach that height? using trace, at 1.25seconds the ball reaches 28ft.

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Hitting objects on the ground Lucy and Ethel are playing yards darts. They are launching the dart 12 ft from the front edge of a circular target of radius 18in on the ground. If Lucy throws the dart directly at the target, and releases it 2.5ft above the ground with an initial velocity of 25ft/sec at a 75° angle, will the dart hit the target? x = (25cos75°)t y = -16t 2 + (25sin75°)t + 2.5 Use Quadratic formula, find when y=0 At about 1.607sec the dart is 10.398 feet from Lucy, which is 12-10.398 = 1.602ft short of the target. How far can the dart travel and be inside the target? 12 to 15 feet.

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Hitting Moving Objects The path of the dart is: x=(125cos15°)t y= -16t 2 + (125sin15°)t +3.5 The path of the monkey is: x = 150 (stays a constant 150 ft from vet) y=-16t 2 + 30 (no initial velocity, since the monkey just lets go of the branch) t:0, 2 x:0, 200 (greater than the distance between vet and monkey) y:0, 40 (greater than the height of the monkey)

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Hitting Moving Objects By how much does the dart miss the monkey? 150=(125cos15°)t t=1.242sec Dart:y= -16(1.242) 2 + (125sin15°)(1.242) +3.5 y = 19.001ft Monkey:y = -16(1.242) 2 + 30 y= 5.319 ft So the dart misses the monkey by 19.001-5.319 =13.682ft.

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Homework Pg 531: 39,40, 47,48

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