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(C) Given the height function: $x(t) = 100t - \frac{25}{2}t^2$. To find the maximum height, we differentiate $x$ with respect to $t$: $\frac{dx}{dt} =

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$200$

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(C) Given the height function: $x(t) = 100t - \frac{25}{2}t^2$. To find the maximum height, we differentiate $x$ with respect to $t$: $\frac{dx}{dt} = 100 - 25t$. For maximum height, set the first derivative to zero: $100 - 25t = 0 \implies t = 4 	ext{ s}$. Now, check the second derivative to confirm it is a maximum: $\frac{d^2x}{dt^2} = -25$. Since $\frac{d^2x}{dt^2} Substitute $t = 4$ into the original equation: $x_{	ext{max}} = 100(4) - \frac{25}{2}(4)^2 = 400 - \frac{25}{2}(16) = 400 - 200 = 200 	ext{ m}$.

$A$ missile is fired from the ground level and rises $x$ meters vertically upwards in $t$ seconds, where $x = 100t - \frac{25}{2}t^2$. The maximum height reached is: (in $\text{ m}$)

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