SOLUTION: In 1993, Joe Carter hit a homerun over the left field wall at the SkyDome in the bottom of the 9th to give the Blue Jays, and Canada, an unprecedented two World Series Championship

Question 1202228: In 1993, Joe Carter hit a homerun over the left field wall at the SkyDome in the bottom of the 9th to give the Blue Jays, and Canada, an unprecedented two World Series Championships in a row!
The function h = -0.001d? + 0.4d + 3, models the height h feet, of Joe's ball as a function of the
stance travelled, d feet, from home plate.
a. How high above the ground did Joe make contact with the ball?
b.
What was the height of the ball as it sailed over the wall 325 feet from home plate?
c. What was the height of the ball when it began to fall back to the ground?
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
The function h = -0.001d? + 0.4d + 3, models the height h feet, of Joe's ball as a function of the stance travelled, d feet, from home plate
a. How high above the ground did Joe make contact with the ball?
b.
What was the height of the ball as it sailed over the wall 325 feet from home plate?
c. What was the height of the ball when it began to fall back to the ground?
First there is a typo in the function h = -0.001d? + 0.4d + 3, what is ?
Since it is an object having projectile motion it may be
h = -0.001d^2 + 0.4d + 3. Let me take this function and work out
h = -0.001d^2 + 0.4d + 3
The term 3 (ft)shows the height at which contact was made with the ball.
What was the height of the ball as it sailed over the wall 325 feet
plug d = 325 in the function
h= -0.001*(325)^2+(0.4)*325 +3
h= 27.375 ft
In the equation ax^2+bx+c=0 x = -b/2a
a=-0.001 b= 0.4plug in
x value = 200
Plug x =200 in the function to get height when it started falling
use calculator

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