Question

15. 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)

Answer 1

To find the maximum height that the pumpkin reaches, we can set the derivative of the equation equal to 0 and solve for x. The derivative of y = 12 + 105x - 16x^2 is y' = 105 - 32x. Setting y' = 0, we get:

0 = 105 - 32x

32x = 105

x = 105/32

The maximum height is reached when x = 105/32 seconds. To find the maximum height, we can plug this value of x into the original equation:

y = 12 + 105(105/32) - 16(105/32)^2

= 12 + 1053.28125 - 163.28125^2

= 12 + 342.8125 - 106.25

= 248.5625

The maximum height that the pumpkin reaches is 248.5625 feet.

To find the number of seconds it takes for the pumpkin to hit the ground (when its height is 0), we can set y = 0 in the original equation and solve for x:

0 = 12 + 105x - 16x^2

16x^2 - 105x + 12 = 0

(4x - 3)(4x - 4) = 0

The solutions to this equation are x = 3/4 and x = 4/4. The second solution is extraneous, since it represents the time at which the pumpkin was launched (when x=0, y would be 12 feet, not 0 feet). Therefore, the number of seconds it takes for the pumpkin to hit the ground is x = 3/4 seconds.