A) The ball in the ground is represented by h(t)=0, that is, the height is equal to 0, the reference level.
Then, we can find for which values of t we have h(t).
We equal h(t) to 0 and calculate t as:
[tex]\begin{gathered} h(t)=-16t^2+64t=0 \\ 64t-16t^2=0 \\ 16t(\frac{64}{16}-t)=0 \\ t_1=0\text{ (first solution)} \\ t_2=\frac{64}{16}=4\text{ (second solution)} \end{gathered}[/tex]
The ball is in the ground at time t=0 (an instant before it is kicked) and then again at time t=4, that is the value we are looking for: the ball landed again in the ground 4 seconds after kicked.
B) The maximum height can be find in two ways:
- By finding the t-value of the vertex, that in this case will be correspond to the maximum height as this is a concave down parabola with only one extreme point.
- Deriving the function and equaling to 0 and finding t.
If we apply the first method, we have:
[tex]\begin{gathered} h(t)=-16t^2+64x=-16(x^2-4x) \\ -16(x^2-4x) \\ -16(x^2-4x+4-4) \\ -16((x-2)^2-4) \\ -16(x-2)^2+64\longrightarrow\text{Vertex:}(2,64) \end{gathered}[/tex]
As the vertex is at time t=2 seconds, the maximum height happens at t=2.
Answer: A) t = 4 seconds B) t = 2 seconds
NOTE for Part B:
If we derive the expression, we get:
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