Question 1192009
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I'll use x in place of t
And y in place of s(t)


Given equation:
y = -4x^2 + 38x + 3


Your teacher must have mixed up feet with meters because this parabolic projectile equation only works with meters (since half g = 9.8 is 4.9 and it looks like s/he is rounding down to the nearest integer for some reason)


The more accurate projectile equation, when dealing with meters only, is:
y = -4.9x^2 + 38x + 3
where 38 is the initial velocity and 3 is the starting height of the ball


Anyway, I'll ignore the physical real world implications since of course this is just a hypothetical problem to get better practice with parabolas.

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Part (a)


Plug in x = 1 to find its y value after 1 second.
y = -4x^2 + 38x + 3
y = -4(1)^2 + 38(1) + 3
y = 37


You have the correct answer.
Nice work.


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Part (b)


For anything of the form
y = ax^2 + bx + c
the vertex (h,k) has its x coordinate defined as
h = -b/(2a)
The k value is determined based on h


From y = -4x^2 + 38x + 3, we have a = -4, b = 38 and c = 3
Let's find h
h = -b/(2a)
h = -38/(2(-4))
h = 4.75
The ball reaches its highest point at 4.75 seconds.


Plug this as the x value into the original equation to find the paired y value of the vertex.
y = -4x^2 + 38x + 3
y = -4(4.75)^2 + 38(4.75) + 3
y = 93.25


The ball reaches the highest point of 93.25 feet.
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