1. ONE DIMENSIONAL MOTION ACCELERATION IN THE Y DIRECTION NOTES

2. STEPS TO SOLVE PROBLEMS WITH A V 0Y 1.Draw a picture 2.Split the problem in 2 parts, (where the object stops and starts to fall back down) 3.Label all that we know and what we want to know 4.Figure out which K equation to use (Use Parabolic graph to help simplify situation) 5.Plug and Chug 6.Write answer and circle it. Check to see if the answer is reasonable

3. PARABOLIC MOTION The height is the same at the same time from V y =0. The Velocities are equal with opposite directions at these Points as well V y =0

4. EXAMPLE # 1 Mobley throws a rock straight up in the air with an initial velocity of 8m/s. a) How high does it go? b) how long is it in the air? V y =0 V 0y =8 m/s Step 3 Label KnowNeed V 0y = 8 m/s∆y=? V fy = 0 m/st=? a y = -9.81 m/s

5. Pick the right equation: For part A) v 2 = v 0 2 + 2a(Δy) For Part B) v = v 0 + aΔt KnowNeed V 0y = 10 m/s∆y=? V fy = 0 m/st=? a y = -9.81 m/s For part A) v 2 = v 0 2 + 2a(Δy)  0= (10 2 )+2(-9.81) Δy  (19.62) Δy = 100  100/19.62 = Δy  Δy = 5.10m For Part B) v = v 0 + aΔt  0 = 10 +(-9.81) Δt  9.81Δt = 10  Δt = 10/9.81  Δt = 1.01s

6. EXAMPLE # 2 You want to throw a ball straight up into the air so that it reaches a height of 3.3 m above the ground. a) If the ball leaves your hand at a height of 1.2 m above the ground, how fast do you need to throw it? b) How long does it take the ball to hit the ground? You may ignore any effects of air resistance throughout this problem.straight up reaches a height of 3.3 m1.2 mhit the ground Step # 1 Draw and Label

7. You want to throw a ball straight up into the air so that it reaches a height of 3.3 m above the ground. a) If the ball leaves your hand at a height of 1.2 m above the ground, how fast do you need to throw it? b) How long does it take the ball to hit the ground? You may ignore any effects of air resistance throughout this problem.straight upreaches a height of 3.3 m1.2 mhit the ground Know Need to KnowPart 1 V fy = 0 m/sV 0y = ? ∆y=3.3m - 1.2mt = ? a y = -9.81 m/s 2Part 2 V 0y = 0 m/st = ? ∆y=3.3m a y = -9.81 m/s 2

8. Know Need to KnowPart 1 V fy = 0 m/sV 0y = ? ∆y=3.3m - 1.2m a y = -9.81 m/s 2Part 2 V 0y = 0 m/st 1 + t 2 = Total Time = ? ∆y=3.3m a y = -9.81 m/s 2 Choose which K Equation to use 1. y = y 0 + v 0 Δt + ½ a(Δt) 2 (relates position and time) 2. v = v 0 + aΔt (relates velocity and time) 3. v 2 = v 0 2 + 2a(Δy) (relates velocity and position) For Part A) I would choose equation 3 to start with. For part B) I would Choose Equation 1

9. PLUG AND CHUG PART A

10. PLUG AND CHUG PART B OPTION 1 So t 1 + t 2 =.82 +.65  1.47 ~ 1.5

11. PLUG AND CHUG PART B OPTION 2

12. CHECK TO SEE IF REASONABLE AND WE UNDERSTOOD PART A v 2 2 = v 1 2 + 2a(y 2 – y 1 ) 0 = v 1 2 + 2(-9.8 m/s 2 )(3.3 m – 1.2 m) v 1 = 6.4 m/s In this problem, you were first asked to find how fast you needed to throw the ball. In other words, you were asked to find the initial velocity of the ball. Because all equations include initial velocity, you could not use that as a criterion to select the best formula. However, you know that the ball reaches a maximum height (v y = 0 at the top of the motion) of 3.3 m, so the best relation to select is the one that relates position and velocity. The speed of 6.4 m/s is about 14 mph, so is a reasonable answer.

13. CHECK TO SEE IF REASONABLE AND WE UNDERSTOOD PART B y 3 = y 1 + v 1 Δt + ½ a(Δt) 2 0 = 1.2 m + (6.4 m/s) Δt + ½ (-9.8 m/s 2 )( Δt) 2 Δt = -0.16 s or +1.5 s Δt = 1.5 s In part b), you were asked to find how long (time) it took the ball to reach the ground (position) and so the position-time equation was used between the initial and final locations. Because the position-time equation includes a ( Δt) 2 term, it will always give two answers for time. The solution that makes physical sense is the positive time, or 1.5 s after the ball leaves your hand.makes physical sense