1. Unit 6 Notes

2. Free-Fall: When an object is flying through the air and the only force acting on it is gravity. In our mini-lab with the cannons, we analyzed the motion of an object in free-fall. During free-fall, acceleration in the x-direction is a x = 0. Acceleration in the y-direction is a y = 9.8m/s 2 downward.

3. The equations of motion for an object in free-fall are as follows: x = v x t + x 0 y = ½ gt 2 + v 0,y t+y 0 v y = v 0,y + gt (v y ) 2 = (v 0,y ) 2 + 2g (∆y) ***These are the same as the kinematic equations from unit 1 but with a x = 0 and a y = g***

4. Example: A ball gets shot straight upward 200m into the air. If it reaches a height of 200m, what was its initial velocity? Solution: In this case, the ball got shot straight upward, so v x = 0. In the y-direction, we will be using the kinematic equations.

5. The only times we have information from are right when the ball launches and right when the ball is at the top of its path. a=9.8m/s 2 v 0 = ? v f = 0 x 0 =0 x=200m

6. Example A ball rolls off a table with an initial velocity of 2m/s. The table is 2.4m tall. How far did the ball travel horizontally before hitting the floor? First solve for the time it takes the ball to fall 2.4m using y position and acceleration due to gravity: y = ½ gt 2 + v 0,y t + y 0 2.4m = ½ * 9.8m/s 2 (t) 2 (v 0,y and y 0 are both 0) t = 0.7s

7. The ball hits the floor after 0.7s and had an initial velocity of 2m/s in the x-direction. How far did the ball travel horizontally? x = v x t + x 0 x = 2m/s * 0.7s + 0 x = 1.4m ***Much of the time, you need to solve for time using y and use that time to solve for x- displacement.