The rate of change in time with respect to height when the object is 125m above the ground is 0.02m/s
Rate of change
Given the function that models the height reached by the object expressed as:
\(t =\sqrt{\dfrac{s}{5} }\)
The rate of change of the function is expressed as:
\(\frac{dt}{ds} = \frac{1}{5} \times \frac{1}{2}*(\frac{s}{5})^{-1/2}\)
Given that s = 125m
Substituting into the function
\(\frac{dt}{ds} = \frac{1}{5} \times \frac{1}{2}*(\frac{125}{5})^{-1/2}\\\frac{dt}{ds} = \frac{1}{5} \times \frac{1}{2}*(25)^{-1/2}\\\frac{dt}{ds} = \frac{1}{5} \times \frac{1}{2} \times \frac{1}{5} \\\frac{dt}{ds} = \frac{1}{50}\)
Hence the rate of change in time with respect to height when the object is 125m above the ground is 0.02m/s
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