h = 5 cm
temperature of water = 10o Celsius
vertical distance = 20 cm
horizontal distance = 11 cm

 

average = -(1.55+1.53+1.88+1.77+1.88)/5 = -1.72o C/s
minimum = -1.72+.20 = -1.52o C/s		maximum = -1.72-.20 = -1.92o C/s

vertical velocity
v2 = u2 + 2*a*s
v2 = (0)2 + 2(-9.80)(.20)
v2 = 3.92
v = -1.98 m/s

time
t = s/v
t = (.20)/(1.98)
t = .10 s

horizontal velocity
v = s/t
v = (.11)/(.10)
v = 1.10 m/s

speed
A2 + B2 = C2
(-1.98)2 + (1.10)2 = C2
5.13 = C2
2.27 m/s


h = 10 cm
temperature of water = 10o Celsius
vertical distance = 20 cm
horizontal distance = 16 cm



average = -(1.78+1.63+1.94+1.94+1.77)/5 = -1.81o C/s
minimum = -1.81+.20 = -1.61o C/s		maximum = -1.81-.20 = -2.01o C/s

vertical velocity
v2 = u2 + 2*a*s
v2 = (0)2 + 2(-9.80)(.20)
v2 = 3.92
v = -1.98 m/s

time
t = s/v
t = (.20)/(1.98)
t = .10 s

horizontal velocity
v = s/t
v = (.16)/(.10)
v = 1.60 m/s

speed
A2 + B2 = C2
(-1.98)2 + (1.60)2 = C2
6.48 = C2
2.55 m/s


h = 15 cm
temperature of water = 10o Celsius
vertical distance = 20 cm
horizontal distance = 18 cm



average = -(1.88+1.75+1.61+1.82+2.00)/5 = -1.81o C/s
minimum = -1.81+.20 = -1.61o C/s		maximum = -1.81-.20 = -2.01o C/s

vertical velocity
v2 = u2 + 2*a*s
v2 = (0)2 + 2(-9.80)(.20)
v2 = 3.92
v = -1.98 m/s

time
t = s/v
t = (.20)/(1.98)
t = .10 s

horizontal velocity
v = s/t
v = (.18)/(.10)
v = 1.80 m/s

speed
A2 + B2 = C2
(-1.98)2 + (1.80)2 = C2
7.16 = C2
2.68 m/s


v = 0 m/s
temperature of water = 10o Celsius



average = -(1.04+1.00+1.33+1.26+1.18)/5 = -1.16o C/s
minimum = (-1.16+.20) = -.96o C/s		maximum = (-1.16-.20) = -1.36o C/s


uncertainty
	
uncertainty of h = 5 cm 
(1.88-1.53)/2 =   .18

uncertainty of h = 10 cm
(1.94-1.63)/2 =   .16

uncertainty of h = 15 cm
(2.00-1.61)/2 =   .20

uncertainty of v = 0 m/s
(1.33-1.00)/2 =   .17

largest uncertainty =   .20