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Algebra 2

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.

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  1. 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)

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  2. 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

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    bobpursley
  3. Max height=184.27ft
    Hits ground after 6.67 seconds

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