Today in Precalculus

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

2. Simulating Projectile Motion Suppose that a baseball is thrown from a point y0 feet above ground level with an initial speed of v0ft/sec at an angle θ with the horizon.

3. Simulating Projectile Motion The path of the object is modeled by the parametric equations: x=(v0cosθ)t y= -16t2 + (v0sinθ)t +y0 Note: The x-component is simply d=rt where r is the horizontal component of v0. The y-component is the velocity equation using the y-component of v0.

4. Example Clark hits a baseball at 3ft above the ground with an initial speed of 150ft/sec at an angle of 18° with the horizontal. Will the ball clear a 20ft fence that is 400ft away? The path of the ball is modeled by the parametric equations: x = (150cos18°)t y = -16t2 +(150sin18°)t + 3 The fence can be graphed using the parametric equations: x = 400 y = 20

5. Example t: 0,3 x: 0,450 y: 0, 80 Approximately how many seconds after the ball is hit does it hit the wall? 400= (150cos18°)t t = 2.804 sec How high up the wall does the ball hit? y=-16(2.804)2 + (150sin18°)(2.804)+3 = 7.178ft

6. Example What happens if the angle is 19°? The ball still doesn’t clear the fence.

7. Example What happens if the angle is 20°? The ball still doesn’t clear the fence.

8. Example What happens if the angle is 21°? The ball just clears the fence.

9. Example What happens if the angle is 22°? The ball goes way over the fence.

10. Practice x = (85cos23°)t y = -16t2 +(85sin23°)t The fence can be graphed using the parametric equations: x = 135 y = 10

11. t: 0,3 x: 0,140 y: 0, 30 How long does it take until the ball passes the cross bar? 135= (85cos23°)t t = 1.725 sec How high is the ball when it passes the cross bar? y=-16(1.725)2 + (65sin23°)(1.725) = 9.681ft

12. Homework worksheet