Answer Happy

1. Considering your value for the % difference in the two values, what can you conclude about the slope of the tangent line drawn at a specific point in time on your Height Versus Time graph and the velocity of the ball at that point in time? Explain any significant difference in your two values.
Height Versus Time 1.200 1.000 0.800 0.600 0.400 +(-0.48 m/s) t +(-9.61 m/s) t 0.200 Tangent line at t = 0.150 s 0.000 Height (m) y=(0,797m) y=(-1.92 m/s) t+ ( 0.905 m) 0.000 0.050 0.100 0.150 Time (sec) Fit Lines 0,200 Tangent 0.00 -0.50 -1.00 -1.50 -2.00 -2.50 -3.00 -3.50 Velocity (m/s) 0.250 Data Set 1 Velocity Versus Time v=(-9.62m/s) t+ (-0.49m/s) Area -0.420 m 0.050 0.100 0.150 • 0.000 Area Time (sec) Show Data 0.200 0.250
400 Examine Velocity +(-0.48 m/s)! +(-9.61 m/s) (² 200 Tang Slope of height graph at time selected: y=(-1.92 m/s) t+ (0.905 m) Velocity at time selected: 000 0.000 0.050 0.150 0% difference: 0.100 Time (sec) Examine Displacement Initial time and height: Final time and height: Displacement during time interval t, to t: Area under velocity-time graph from t; to t₁: % difference: Examine Acceleration Slope of velocity graph: Local gravitational acceleration: % difference: Report Time selected: -2.50 -3.00 -3.50 0.150 sec -1.92 m/s v (t₁) = -2.00 m/s % diff= -4.08 Time (sec) 2% y₁=0.800 Y₁ = 0.600 Ay = -2.00 A= -0.420 m % diff=130.57 % v' = -9.62 m/s² g= -9.80 m/s² t₁ = v = (-9.62 my' (t₁) = Area = -0.420 m 0.050 0.000 t = -0.70 sec t= -2.00 sec E E E % diff= -1.85 % 0.25