Answer:
[tex]h(t)=-16t^2+70[/tex]
The time to reach ground is 2.09165 sec
Step-by-step explanation:
(a)
we are given
[tex]h(t)=-16t^2+h_0[/tex]
we are given
height of the building is 70ft
so, at t=0 , h(t)=70
so, we can use it
[tex]h(0)=-16(0)^2+h_0[/tex]
[tex]70=-16(0)^2+h_0[/tex]
[tex]h_0=70[/tex]
now, we can plug it back
[tex]h(t)=-16t^2+70[/tex]
On ground , height becomes 0
so, we can set h(t)=0
and then we can solve for t
[tex]h(t)=-16t^2+70=0[/tex]
[tex]t^2=\frac{35}{8}[/tex]
[tex]t=\frac{\sqrt{70}}{4},\:t=-\frac{\sqrt{70}}{4}[/tex]
Since, we can only consider positive value
and we get
[tex]t=2.09165[/tex]
(b)
we can draw graph
Answer:
It's important to notice that the person is on the top of the building, so its position is the same as the height of the building.
Additionally, the movement is defined by
[tex]h(t)=-16t^{2}+h_{0}[/tex]
Where [tex]h(t)[/tex] is the height at any moment, and [tex]h_{0}[/tex] is the initial height (the same as the building's).
So, we know the building is 70 feet tall, that means [tex]h_{0} =70[/tex].
When the tool hits the ground its height is zero, so [tex]h(t)=0[/tex].
Replacing these values, we have
[tex]h(t)=-16t^{2}+h_{0}\\0=-16t^{2}+70[/tex]
Solving for [tex]t[/tex]
[tex]16t^{2}=70\\ t^{2}=\frac{70}{16}=4.375\\ t=\sqrt{4.375}\\ t \approx 2.09[/tex]
So, it took 2.09 seconds to hit the ground.
On the other hand, to dra a graph that represents the movement of the tool, we just need to graph the quadratic function we used.
The image attached shows the graph of this situation.
Also, the falling red curve shows the height at any second.
