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

2. 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.

3. 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.

4. 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)

5. 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.

6. Homework Pg 531: 39,40, 47,48