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Overview In this experiment we will messure the transfer of kinetic energy to po

ID: 1770159 • Letter: O

Question

Overview In this experiment we will messure the transfer of kinetic energy to potential energy of a projectile launched vertically upwards. This will require us to tueasure the msuzzle velocity of the projectile and it's maximum height above the muzzle. We will neglect air resistance and other possible sources of non-conservative energy loss (or work). Since all of the projectile's kinetic energy will be directed into it's vertical motion, than all of the kinetic energy should be transferred to potential energy at it maximum altitude. Thus we may compare the average kinetic energy of the projectile when it leaves the muzzle to the potential energy stored in the projectile at its maximum height, where it's motion monentarily stops. Theory

Explanation / Answer

Mean of the five velocity= Vavg = 0.0128

dV

V-Vavg

(V-Vavg)2

0.031

0.003

0.000009

0.002

-0.026

0.000676

0.008

-0.02

0.0004

0.015

-0.013

0.000169

0.008

-0.02

0.0004

0.001654

So, Standard deviation = 0.001654

Standard error = SD/(5) = 0.00074

Margin of error = 2X0.00074 = 0.0015

So, Confidence interval = 0.0128-0.0015 = 0.0113

                                And, 0.0128+0.0015 = 0.0143

So the 95% confidence interval is, 0.0113 to 0.0143,

So the data seems not like to confident.

Mean of the five velocity= Vavg = 0.0128

dV

V-Vavg

(V-Vavg)2

0.031

0.003

0.000009

0.002

-0.026

0.000676

0.008

-0.02

0.0004

0.015

-0.013

0.000169

0.008

-0.02

0.0004

0.001654

So, Standard deviation = 0.001654

Standard error = SD/(5) = 0.00074

Margin of error = 2X0.00074 = 0.0015

So, Confidence interval = 0.0128-0.0015 = 0.0113

                                And, 0.0128+0.0015 = 0.0143

So the 95% confidence interval is, 0.0113 to 0.0143,

So the data seems not like to confident.