## Precalculus (6th Edition) Blitzer

The velocity of the ball after 2 seconds is approximately $-24$ feet per second.
Consider the details; the throwing function is $s\left( t \right)=-16{{t}^{2}}+40t$ after $t$ seconds. Now, the velocity is: \begin{align} & {s}'\left( t \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( \frac{s\left( t+h \right)-s\left( t \right)}{h} \right) \\ & =\underset{h\to 0}{\mathop{\lim }}\,\left( \frac{-16{{\left( t+h \right)}^{2}}+40\left( t+h \right)-\left( -16{{t}^{2}}+40t \right)}{h} \right) \\ & =\underset{h\to 0}{\mathop{\lim }}\,\left( \frac{-16{{t}^{2}}-32ht-16{{h}^{2}}+40t+40h+16{{t}^{2}}-40t}{h} \right) \\ & =\underset{h\to 0}{\mathop{\lim }}\,\left( \frac{-32ht-16{{h}^{2}}+40h}{h} \right) \end{align} This implies, \begin{align} & {s}'\left( t \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( \frac{h\left( -32t-16h+40 \right)}{h} \right) \\ & =\underset{h\to 0}{\mathop{\lim }}\,\left( -32t-16h+40 \right) \\ & =-32t-16\left( 0 \right)+40 \\ & =-32t+40 \end{align} At 2 seconds, substitute $t=2$ in the velocity equation ${s}'\left( t \right)=-32t+40$. \begin{align} & {s}'\left( t \right)=-32t+40 \\ & {s}'\left( 2 \right)=-32\left( 2 \right)+40 \\ & =-64+40 \\ & =-24 \end{align} Hence, the instantaneous velocity of the ball after 2 seconds is approximately $-24$ feet per second.