Aball is launched into the sky at 272 feet per second from the roof of a skyscraper 1,344 feet tall. the equation for the ball’s height h at time t seconds is h = -16t2 + 272t + 1344. when will the ball strike the ground?

When the ball is in the ground, its height is basically equal to zero thus making our equation,

    -16t^2 + 272t + 1344 = 0

Simplifying the equation will give us,

  - t^2 + 17t + 84 = 0 or t^2 – 17t – 84 = 0

Factoring out the equation will give us,

   (t – 21)(t + 4) = 0

Thus, t = 21 or t = -4. -4 is an extraneous root. Thus, the answer is t = 21.

21 seconds

x = 25/2304

step-by-step explanation:

answer: r there any options?

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