To solve this problem, we will start by rewriting the function \(h(t) = -3t^2 + 12t + 96\) and then find when the drone lands on the ground by solving \(h(t) = 0\).
1) Rewriting the function
We can factor the equation by first factoring out the coefficient of \(t^2\) from the quadratic.
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Factor out \(-3\): \[ h(t) = -3(t^2 - 4t - 32) \]
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Now we will factor the quadratic \(t^2 - 4t - 32\). To factor this, we need to find two numbers that multiply to \(-32\) (the constant term) and add to \(-4\) (the coefficient of \(t\)). The factors are \(-8\) and \(4\).
So, we can write: \[ t^2 - 4t - 32 = (t - 8)(t + 4) \]
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Incorporating the factor of \(-3\), we rewrite \(h(t)\): \[ h(t) = -3(t - 8)(t + 4) \]
Thus, the function in its factored form is: \[ h(t) = -3(t - 8)(t + 4) \]
2) Finding when the drone lands on the ground
To determine when the drone lands on the ground, we set \(h(t)\) to 0: \[ -3(t - 8)(t + 4) = 0 \]
This will give us two solutions: \[ (t - 8) = 0 \quad \text{or} \quad (t + 4) = 0 \]
From the first equation: \[ t - 8 = 0 \Rightarrow t = 8 \]
From the second equation: \[ t + 4 = 0 \Rightarrow t = -4 \]
Since time cannot be negative in this context, we discard \(t = -4\).
Thus, the drone lands on the ground \(8\) minutes after takeoff.
Final Answers
- \[h(t) = -3(t - 8)(t + 4)\]
- The drone lands on the ground after \(8\) minutes.