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Home » Using the equation \( v^2 = u^2 - 2gh \): - AssociationVoting

Using the equation \( v^2 = u^2 - 2gh \): - AssociationVoting

📅 May 15, 2026 👤 scraface

May 15, 2026

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\[ u' = k, \quad v' = 2t \]
\[ C'(t) = \frac{(k)(t^2 + 1) - (kt)(2t)}{(t^2 + 1)^2} \]
\[ C'(t) = \frac{k(t^2 + 1) - 2kt^2}{(t^2 + 1)^2} \]
\[ C'(t) = \frac{kt^2 + k - 2kt^2}{(t^2 + 1)^2} \]
\[ C'(t) = \frac{-kt^2 + k}{(t^2 + 1)^2} \]
\[ C'(t) = \frac{k(1 - t^2)}{(t^2 + 1)^2} \]
Set \(C'(t) = 0\):
\[ \frac{k(1 - t^2)}{(t^2 + 1)^2} = 0 \]
The numerator must be zero:
\[ 1 - t^2 = 0 \]

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