Answer Happy

Considering your value for the % difference in the two values, what can you conclude about the area under your Velocity Versus Time graph for the selected interval of time and the displacement of the ball during that same interval of time? Explain any significant difference in your two values.
Height Versus Time Height (m) 1.200 1.000 0.800 0.600 0.400 y=(0,797m) +(-0.48 m/s) +(-9.61 m/s) t 0.200 Tangent line at t= 0.150 s 0.000 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=(-9.62 m/s) t+ (-0.49m/s) Area -0.420 m 0.050 • 0.000 Area 0.100 0.150 0.200 Time (sec) Show Data 0.250

400 Examine Velocity +(-0.48 m/s) t +½ (-9.61 m/s) t² 200 Tan 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 t₁ = v = (-9.62 my' (t₁) = Area = -0.420 m 0.050 0.150 -1.92 v (t₁) = -2.00 % diff= -4.08 Time (sec) Y₁ = 0.800 y₁ = 0.600 m Ay = -2.00 A= -0.420 m % diff=130.57 % v' = -9.62 m/s² g= -9.80 m/s² -1.85 % 0.000 t = -0.70 t= -2.00 sec sec % diff= sec m/s m/s 2% E E 0.25