Answer
The rocket climbs at a speed of $60ft/s$ at $10$ sec after liftoff.
Work Step by Step
The function of the height of the rocket after $t$ seconds of liftoff is $f(t)=3t^2$ ft.
The velocity of the rocket at $t=10$ seconds, therefore, is the rate of change of the height $f(t)$ with respect to $t=10$, or in other words, the derivative $f'(10)$, which can be calculated by $$f'(10)=\lim_{h\to0}\frac{f(h+10)-f(10)}{h}=\lim_{h\to0}\frac{3(h+10)^2-(3\times10^2)}{h}$$
$$f'(10)=\lim_{h\to0}\frac{3(h^2+20h+100)-300}{h}=\lim_{h\to0}\frac{3h^2+60h+300-300}{h}$$
$$f'(10)=\lim_{h\to0}\frac{3h^2+60h}{h}=\lim_{h\to0}(3h+60)$$
$$f'(10)=3\times0+60=60(ft/s)$$
The rocket climbs at a speed of $60ft/s$ at $10$ sec after liftoff.
