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A ball falls freely from rest with an acceleration g. The variation with time t of its displacement s is given by s=1/2gt^2. What is the percentage uncertainty in the value of s if the percentage uncertainties in t and g are ±3% and ±2% respectively?

A. ±6%
B. ±5%
C. ±8%
D. ±1%

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Final answer:

The percentage uncertainty in the displacement s of a freely falling ball can be found by combining the uncertainties in time t and acceleration g. The calculation involves doubling the percentage uncertainty of the squared time (6%) and adding it to the uncertainty in g (2%), resulting in a total uncertainty of ±8%.

Step-by-step explanation:

To calculate the percentage uncertainty in the displacement s when a ball falls freely with acceleration g and the variation with time t of its displacement is given by the equation s = 1/2gt^2, we need to use the given percentage uncertainties in t and g. We know that the percentage uncertainties in t and g are ±3% and ±2% respectively.

Since the equation for displacement is a product of t squared and g, we can use the rule for combining uncertainties for multiplication and powers. The percentage uncertainty for a variable raised to a power is the percentage uncertainty in that variable times the power. For the time t squared that would be 2 * 3% = 6%. Since the uncertainty in g is 2%, the total percentage uncertainty in s is the sum of the uncertainties in g and t squared, which is 6% + 2% = ±8%. Therefore, the correct answer is C. ±8%.

User Toni Villena
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