Consider right triangle with vertices B - base of the hill, S - top of the statue and Y - you. In this triangle angle B is right and angle Y is 13.2°. If h is a height of the statue, then the legs YB and BS have lengths 77 ft and 16+h ft.
You have lengths of two legs and measure of one acute angle, then you can use tangent to find h:
ft.
Answer: the height of the statue is 2.0565 ft.
From the given function modeling the height of the ball:
f(x)=-0.2x^2+1.4x+7
A] The maximum height of the ball will be given by:
At max height f'(x)=0
from f(x),
f'(x)=-0.4x+1.4
solving for x we get:
-0.4x=-1.4
x=3.5ft
thus the maximum height would be:
f(3.5)=-0.2(3.5)^2+1.4(3.5)+7
f(3.5)=9.45 ft
b]
How far from where the ball was thrown did this occur:
from (a), we see that at maximum height f'(x)=0
f'(x)=-0.4x+1.4
solving for x we get:
-0.4x=-1.4
x=3.5ft
This implies that it occurred 3.5 ft from where the ball was thrown.
c] How far does the ball travel horizontally?
f(x)=-0.2x^2+1.4x+7
evaluationg the expression when f(x)=0 we get:
0=-0.2x^2+1.4x+7
Using quadratic equation formula:
x=-3.37386 or x=10.3739
We leave out the negative and take the positive answer. Hence the answer 10.3739 ft horizontally.
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