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 – 16x^2 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.)
A.
The pumpkin's maximum height is 184.27 feet and it hits the ground after 3.28 seconds.
B.
The pumpkin's maximum height is 3.28 feet and it hits the ground after 6.67 seconds.
C.
The pumpkin's maximum height is 184.27 feet and it hits the ground after 6.67 seconds.
D.
The pumpkin's maximum height is 3.28 feet and it hits the ground after 184.27 seconds.
y = 12 + 105t – 16t^2 so when is height equal to zero?
0 = 12 + 105t – 16t^2
t= (-105+-sqrt(105^2+4*16*12))/-32=> t=6.67 sec check that
max height is when t=6.67/2=3.335
height=12 + 105*3.335– 16*3.335^2=about 184 feet
Max height=184.27ft
Hits ground after 6.67 seconds
To find the maximum height that the pumpkin reaches, we need to determine the vertex of the quadratic equation y = 12 + 105x – 16x^2.
The vertex of a quadratic equation in the form y = ax^2 + bx + c is given by the formula x = -b / (2a).
In this case, a = -16 and b = 105.
Substituting these values into the formula, we have x = -105 / (2*(-16)) = 105 / 32.
To find the maximum height, we substitute this value of x back into the equation:
y = 12 + 105(105/32) - 16(105/32)^2.
Evaluating this expression, we find that the maximum height is approximately 184.27 feet. (Option A is correct.)
Now, let's find the time it takes for the pumpkin to hit the ground. Since the height is 0 feet when the pumpkin hits the ground, we need to solve the equation:
0 = 12 + 105x – 16x^2.
This is a quadratic equation, so we can solve it by factoring, completing the square, or using the quadratic formula. In this case, let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a).
Plugging in the values a = -16, b = 105, and c = 12, we have:
x = (-105 ± √(105^2 - 4*(-16)*12)) / (2*(-16)).
Calculating this expression, we find two possible solutions: x ≈ 3.28 and x ≈ 6.67.
Since the time cannot be negative, we discard the negative solution, and thus the pumpkin hits the ground after approximately 3.28 seconds. (Option A is correct.)
nothing hard here.
As with all y=ax^2+bx+c,
the vertex (max height) is at (-b/2a, c - b^2/4a)
and you can use the quadratic formula to solve for the roots (when the height is zero)