Question

A ball is thrown upward and outward from a height of 77 feet. the height of the​ ball, f(x), in​ feet, can be modeled by f left parenthesis x right parenthesis equals negative 0.2 x squared plus 1.4 x plus 7f(x)=−0.2x2+1.4x+7 where x is the​ ball's horizontal​ distance, in​ feet, from where it was thrown. use this model to solve parts​ (a) through​ (c).

Answer 1

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.