(B) Given the height function: $x = 80t - 16t^2$. To find the time at which the stone reaches maximum height,we differentiate $x$ with respect to $t$:

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(B) Given the height function: $x = 80t - 16t^2$. To find the time at which the stone reaches maximum height,we differentiate $x$ with respect to $t$: $\frac{dx}{dt} = 80 - 32t$. At maximum height,the velocity of the stone is zero,so $\frac{dx}{dt} = 0$. Setting the derivative to zero: $80 - 32t = 0$. $32t = 80$. $t = \frac{80}{32} = 2.5 \text{ s}$.

Class 12 Mathematics Application of Derivatives Maxima and Minima

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$A$ stone is thrown vertically upwards and the height $x \text{ ft}$ reached by the stone in $t$ seconds is given by $x = 80t - 16t^2$. The stone reaches the maximum height in (in $\text{ s}$)

KCET 2008 All KCET

A
$2$
B
$2.5$
C
$3$
D
$1.5$

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Application of Derivatives

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Maxima and Minima

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$A$ stone is thrown vertically upwards and the height $x \text{ ft}$ reached by the stone in $t$ seconds is given by $x = 80t - 16t^2$. The stone reaches the maximum height in (in $\text{ s}$)

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