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 = 10 + 95x – 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.)
To find the maximum height that the pumpkin reaches, we need to find the vertex of the quadratic equation. The equation is in the form of y = ax^2 + bx + c, where a = -16, b = 95, and c = 10.
The x-coordinate of the vertex can be found using the formula x = -b / (2a).
x = -95 / (2 * -16) = -95 / -32 ≈ 2.97
To find the y-coordinate of the vertex, we substitute the x-coordinate into the equation.
y = 10 + 95(2.97) - 16(2.97)^2 ≈ 146.37
Therefore, the maximum height that the pumpkin reaches is approximately 146.37 feet.
To find the time when the pumpkin hits the ground, we set the height equation equal to 0 and solve for x.
0 = 10 + 95x - 16x^2
Rearranging the equation, we get:
16x^2 - 95x - 10 = 0
We can solve this quadratic equation using factoring or the quadratic formula. However, since the equation is not easily factorable, we will use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values, we have:
x = (-(-95) ± √((-95)^2 - 4 * 16 * -10)) / (2 * 16)
x = (95 ± √(9025 + 640)) / 32
x = (95 ± √9665) / 32
Since the problem asks for the time when the pumpkin hits the ground, we only need the positive solution:
x ≈ (95 + √9665) / 32 ≈ 6.62
Therefore, approximately 6.62 seconds have passed when the pumpkin hits the ground.