A potato was launched in the air using a potato gun. the function for the situation was h(t) = -16t2 + 100t + 10, where t was the time in seconds and h(t) was the height in feet. (3 points each) a. what was the maximum height of the potato? b. what was the lowest height of the potato? c. what was the lowest time that applies to the actual situation? d. what is the greatest time that applied to the actual situation? e. using your answers to parts a through d, what are the domain and range of the function, as they apply to this situation? use set builder notation.

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tutorconsortium33 tutorconsortium33 · Answer 1 · 2022-08-01T15:27:04+02:00

The answers to all the parts are given below.

The height equation is:

h(t) = - 16t^2 + 100t + 10

A) To find the maximum height, we must find the time where the velocity is 0, so we must derive it with respect to the time.

v(t) = - 2 * 16 * t + 100 = 0

t = 100/(2*16) = 3.125s

Now we put this time in our height equation:

h(3.125s) = -16*3.125^2 + 100*3.125 + 10 = 166.25 ft.

B) The lowest height of the potato will be h = 0ft when the potato hits the ground.

C) the lowest time that works for that function is t = 0s when the potato is fired by the gun.

D) the maximum time can be found when the potato hits the ground, after that point the equation does not work anymore, let's find it,

h(t) = 0 = -16t^2 +100t+10

we can solve it using Bhaskara's equation:

[tex]t=\frac{-100+-\sqrt{100^{2} +4*16*10} }{-2*16} =\frac{-100+-103.2}{-32}[/tex]

So the two solutions are:

t = 6.3s

t = -0.1s

We need to choose the positive time, as we already discussed that the minimum time that works for the equation is t = 0s.

so here the answer is t = 6.3s

E) the domain is 0s ≤ t ≤ 6.3s

The range is 0ft ≤ h ≤ 166.25 ft.

Therefore, the answers to all the parts are shown.

Know more about a function here:

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