Every year in Delaware there is a contest where people create cannons and catapults designed to launch pumpkins as far in the air as possible. The equation y=12+105x−16x2 can be used to represent the height, y, of a launched pumpkin, where x is the time in seconds that the pumpkin has been in the air. What is the maximum height that the pumpkin reaches? How many seconds have passed when the pumpkin hits the ground? (Hint: If the pumpkin hits the ground, its height is 0 feet.)(1 point)
To find the maximum height that the pumpkin reaches, we need to find the vertex of the parabola described by the equation. The x-coordinate of the vertex can be found using the formula x = -b/2a, where a = -16 and b = 105 in this case.
x = -105 / (2*(-16))
x = -105 / (-32)
x ≈ 3.28
To find the y-coordinate of the vertex, we substitute this value of x back into the equation:
y = 12 + 105(3.28) - 16(3.28)^2
y ≈ 12 + 344.4 - 169.976
y ≈ 186.424
Therefore, the maximum height that the pumpkin reaches is approximately 186.424 feet.
To find the time when the pumpkin hits the ground, we set the height equation to 0 and solve for x:
0 = 12 + 105x - 16x^2
16x^2 - 105x - 12 = 0
We can then solve this quadratic equation using factoring, the quadratic formula, or completing the square. Using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
x = (-(-105) ± √((-105)^2 - 4(16)(-12))) / (2(16))
x = (105 ± √(11025 + 768)) / 32
x = (105 ± √11793) / 32
Since the problem asks for the time when the pumpkin hits the ground, we can ignore the negative solution:
x = (105 + √11793) / 32
x ≈ 6.241
Therefore, approximately 6.241 seconds have passed when the pumpkin hits the ground.