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A stone is thrown vertically upwards and the height $x f$ t. reached by the stone in $t$ sec is given by $x=80 t-16 t^{2}$. The stone reaches the maximum height in
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Verified Answer
The correct answer is:
$2.5 \mathrm{sec}$
We have,
$$
\begin{aligned}
x &=80 t-16 t^{2} \\
\Rightarrow \quad \frac{d x}{d i} &=80-32 t
\end{aligned}
$$
For maximum value of $x$,
$$
\begin{aligned}
\frac{d x}{d t} &=0 \\
\Rightarrow 80-32 t &=0 \Rightarrow t=\frac{80}{32}=2.5
\end{aligned}
$$
Again, $\frac{d^{2} x}{d t^{2}}=-32 < 0$
So, at $t=25 \mathrm{sec}, x$ is maximum.
$$
\begin{aligned}
x &=80 t-16 t^{2} \\
\Rightarrow \quad \frac{d x}{d i} &=80-32 t
\end{aligned}
$$
For maximum value of $x$,
$$
\begin{aligned}
\frac{d x}{d t} &=0 \\
\Rightarrow 80-32 t &=0 \Rightarrow t=\frac{80}{32}=2.5
\end{aligned}
$$
Again, $\frac{d^{2} x}{d t^{2}}=-32 < 0$
So, at $t=25 \mathrm{sec}, x$ is maximum.
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