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Naily [24]
3 years ago
8

The equation g(t)=2+23.7t−4.9t2 models the height, in meters, of a pumpkin t seconds after it has been launched from a catapult.

Is the pumpkin still in the air 8 seconds later? Explain or show how you know.
Mathematics
1 answer:
inysia [295]3 years ago
5 0

Answer:

The pumpkin is not in the air

Step-by-step explanation:

The equation is

g(t)=2+23.7t-4.9t^2

Let us find when the pumpkin will reach the ground

0=2+23.7t-4.9t^2\\\Rightarrow -49t^2+237t+20=0\\\Rightarrow t=\frac{-237\pm \sqrt{237^2-4\left(-49\right)\times 20}}{2\left(-49\right)}\\\Rightarrow t=-0.08,4.91

So, the pumpkin will reach the ground at 4.91 seconds. Hence, at 8 seconds the pumpkin will reach the ground.

<h2>OR</h2>

At t=8\ \text{s}

g(8)=2+23.7\times 8-4.9\times 8^2\\\Rightarrow g(8)=-122\ \text{m}

The height of the pumpkin is negative 122 m. This means it is lower than the ground. So, the pumpkin is not in the air 8 seconds later.

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