Answer
(a) $56.34$ m/s upward
(b) $v(a)=58-1.66a$ m/s
(c) $69.88$ seconds
(d) $-58$ m/s downward.
Work Step by Step
(a) Given $H=58t-0.83t^2$, as velocity is the change in height per unit time, we have:
$$v(t)=H'(t)=\lim_{h\to 0}\frac{H(t+h)-H(t)}{h}=\lim_{h\to 0}\frac{58(t+h)-0.83(t+h)^2-(58t-0.83t^2)}{h}=\lim_{h\to 0}\frac{58h-0.83h^2-0.83\times2ht}{h}=\lim_{h\to 0}(58-0.83h-1.66t)=58-1.66t$$
Thus we have $v(1)=58-1.66=56.34$ m/s upward
(b) When $t=a$, we have $v(a)=58-1.66a$ m/s
(c) The arrow will hit the moon when $H(t)=0, t\ne0$, we have $H=58t-0.83t^2=0$ and $t=\frac{58}{0.83}\approx69.88$ seconds
(d) At $t=69.88$, we have $v=58-1.66\times59.88=-58$ m/s downward.
