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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
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The correct answer is:
$2.5 \mathrm{~s}$
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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