Question

The height, s of a ball (in feet) thrown with an initial velocity of 80 feet per second from an initial height of 6 feet is given as a function of the time t (in seconds) by s(t)=-16t^2 +80t+6. Explain in detail how you could find the maximum height of the ball. Now, explain, in detail, a second method of finding the maximum height. Using one of these two methods, determine the time at which height is a maximum and find that maximum height. Be sure to use units in your answer and explain your methods thoroughly.

Answer 1

`\begin{gathered} \text{ One way of solving the maximum height of the ball with the given equation is} \\ \text{getting the vertex of the function of time since it is a quadratic function given by} \\ x=-(b)/(2a),\text{ and substitute time t to the original function} \\ \text{ The other method of finding the maximum height is getting} \\ \text{ the first derivative of the function},\text{ and equate it to first derivative to zero} \\ \text{Solving the maximum height using vertex of the function} \\ \text{ Convert }x=-(b)/(2a)\text{ to a function of time we get} \\ t=-(80)/(2(-16))=-(80)/(-32) \\ t=2.5\text{seconds} \\ \text{Substitute} \\ s(t)=-16t^2+80t+6 \\ s(2.5)=-16(2.5)^2+80(2.5)+6 \\ s(2.5)=106ft \\ \text{Solving using the first derivative of the function} \\ s(t)=-16t^2+80t+6 \\ s^(\prime)(t)=(2)(-16)t+80+0 \\ s^(\prime)(t)=-32t+80 \\ -32t+80=0 \\ -32t=-80 \\ (-32t)/(-32)=(-80)/(-32) \\ t=2.5\text{seconds} \\ \text{ Substitute as usual and we get the same result.} \end{gathered}`