KCET · Maths · Parabola
A stone is thrown vertically upwards and the height \(x \mathrm{ft}\) reached by the stone in t seconds is given by \(x=80 t-16 t^{2}\). The stone reaches the maximum height in
- A \(2 \mathrm{~s}\)
- B \(2.5 \mathrm{~s}\)
- C \(3 \mathrm{~s}\)
- D \(1.5 \mathrm{~s}\)
Answer & Solution
Correct Answer
(B) \(2.5 \mathrm{~s}\)
Step-by-step Solution
Detailed explanation
Given, \(x=80 t-16 t^{2}\)
On differentiating w.r.t. , we get
\[
\frac{\mathrm{dx}}{\mathrm{dt}}=80-32 \mathrm{t}
\]
Since, the stone reaches the maximum height, then velocity will be zero \(\left(\right.\) ie, \(\left.\frac{\mathrm{dx}}{\mathrm{dt}}=0\right)\)
\(\therefore\) \(\Rightarrow\) Let \(y=\frac{\log x}{x}\)
\(80-32 t=0\)
\(t=2.5 \mathrm{~s}\)
On differentiating w.r.t. , we get
\[
\frac{\mathrm{dx}}{\mathrm{dt}}=80-32 \mathrm{t}
\]
Since, the stone reaches the maximum height, then velocity will be zero \(\left(\right.\) ie, \(\left.\frac{\mathrm{dx}}{\mathrm{dt}}=0\right)\)
\(\therefore\) \(\Rightarrow\) Let \(y=\frac{\log x}{x}\)
\(80-32 t=0\)
\(t=2.5 \mathrm{~s}\)
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