Given the equation of the path of the football: y = − 16/1225x^2 + 7/5x + 3/2
(a) How high is the ball when it is punted?
We can find y when x = 0 by substituting x = 0 in the equation:
y = − 16/1225x^2 + 7/5x + 3/2
y = − 16/1225 * 0^2 + 7/5 * 0 + 3/2
y = 3/2
So, the ball is 3/2 feet high when it is punted.
(b) What is the maximum height of the football?
To find the maximum height, we can differentiate the equation and set it equal to 0, then solve for x.
y' = −32/1225x + 7/5
0 = −32/1225x + 7/5
x = 7/32/1225
Then, substitute x back into the equation to find y:
y = − 16/1225x^2 + 7/5x + 3/2
y = − 16/1225 * (7/32/1225)^2 + 7/5 * (7/32/1225) + 3/2
y = −16/1225 * 49/32/1225 + 7/5 * 7/32/1225 + 3/2
y = −16/1225 * 49/32 + 7/5 * 7/32 + 3/2
y = −49/32 + 7/5 * 7/32 + 3/2
y = 49/40 - 49/32 + 3/2
y = 49/40 + 3/2 - 49/32
y = 49/40 + 3/2 - 49/32
y = 4.86 - 1.53
y = 3.33
The maximum height of the football is 3.33 feet.
(c) How far from the punter does the football strike the ground?
To find the x-coordinate where the football hits the ground, we need to find when y = 0:
y = − 16/1225x^2 + 7/5x + 3/2
0 = − 16/1225x^2 + 7/5x + 3/2
−3/2 = − 16/1225x^2 + 7/5x
16/1225x^2 = 7/5x + 3/2
1225/16 * x^2 = 15/7 * x + 15/2
x^2 = 15/7 * x/1225 * 16 + 15/2 * 1225/16
x^2 = 15/7 * x/1225 * 16 + 15 * 50
x^2 = 15/7 * x/1225 * 16 + 750
x^2 - 750 = 15/7 * x/1225 * 16
x^2 - 750 = 15/7 * x * 16/1225
x^2 = 750 + 15/7 * x * 16/1225
x^2 = 750 + 120/7 * x/1225
x = ±√(750 + 120/7 * x/1225)
We need to find the positive root, since x represents the horizontal distance from the punter.
x = √(750 + 120/7 * x/1225)