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A stone is thrown vertically upwards from the top of a tower $64 \mathrm{~m}$ high according to the law $s=48 t-16 t^{2}$. The greatest height attained by the stone above ground is
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The correct answer is:
$100 \mathrm{~m}$
Given, $s=48 t-16 t^{2}$
$\Rightarrow \quad \frac{\mathrm{ds}}{\mathrm{dt}}=48-32 \mathrm{t}$
At the greatest height, $\mathrm{v}=\frac{\mathrm{ds}}{\mathrm{dt}}=0$
$\Rightarrow \quad 48-32 t=0$
$\Rightarrow \quad \mathrm{t}=\frac{3}{2}$
$\therefore \quad \mathrm{s}=48 \times \frac{3}{2}-16\left(\frac{3}{2}\right)^{2}$
$=36 \mathrm{~m}$
$\therefore$ Total height $=64+36=100 \mathrm{~m}$
$\Rightarrow \quad \frac{\mathrm{ds}}{\mathrm{dt}}=48-32 \mathrm{t}$
At the greatest height, $\mathrm{v}=\frac{\mathrm{ds}}{\mathrm{dt}}=0$
$\Rightarrow \quad 48-32 t=0$
$\Rightarrow \quad \mathrm{t}=\frac{3}{2}$
$\therefore \quad \mathrm{s}=48 \times \frac{3}{2}-16\left(\frac{3}{2}\right)^{2}$
$=36 \mathrm{~m}$
$\therefore$ Total height $=64+36=100 \mathrm{~m}$
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