## Calculus: Early Transcendentals 8th Edition

(a) The velocity of the rock after 1 second is $v_a=6.28m/s$ (b) The velocity of the rock after a seconds is $v_b=10-3.72a \hspace{0.1cm}(m/s)$ (c) The rock would hit the ground after about $5.376s$ (d) The velocity of the rock as it hits the ground is $v_d\approx-10m/s$
The function of height after $t$ seconds: $$H=f(t)=10t-1.86t^2$$ The velocity of a rock thrown into the air: $v_i=10m/s$ *Find the function of the change of velocity over time: According to definition, the function of the instantaneous $v(a)$ at time $t=a$ is $$v(a)=\lim\limits_{h\to0}\frac{f(t+h)-f(t)}{h}$$$$v(a)=\lim\limits_{h\to0}\frac{10(t+h)-1.86(t+h)^2-(10t-1.86t^2)}{h}$$$$v(a)=\lim\limits_{h\to0}\frac{(10t+10h)+(-1.86t^2-1.86h^2-3.72th)-10t+1.86t^2}{h}$$$$v(a)=\lim\limits_{h\to0}\frac{10h-1.86h^2-3.72th}{h}$$$$v(a)=\lim\limits_{h\to0}(10-1.86h-3.72t)$$$$v(a)=10-1.86\times0-3.72t$$$$v(a)=10-3.72t$$ (a) The velocity of the rock after 1 seconds $(t=1)$ is $$v_a=10-3.72\times1=6.28m/s$$ (b) The velocity of the rock after a seconds $(t=a)$ is $$v_b=10-3.72a \hspace{0.1cm}(m/s)$$ (c) The rock will hit the surface when its height $H=0$. That means $$H=10t-1.86t^2=0$$$$t(10-1.86t)=0$$$$t=0s \hspace{0.5cm} or \hspace{0.5cm}t\approx5.376s$$ $t=0s$ corresponds to the time the rock was thrown in the air. Therefore, it is not counted. So, the rock would hit the ground after about $5.376s$. (d) The velocity of the rock as it hits the ground $(t=5.376s)$ is $$v_d=10-3.72\times5.376\approx-10m/s$$