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A rocket is launched from a tower. The height of the rocket,y in feet, is related to the time after launch, x in seconds, by the given equation. Using this equation, find out the time at which the rocket will reach its max, to the nearest 100th of a second. y=-16x^2+171x+87

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A rocket is launched from a tower. The height of the rocket,y in feet, is related to the time after launch, x in seconds, by the given equation. Using this equation, find out the time at which the rocket will reach its max, to the nearest 100th of a second. y=-16x^2+171x+87

A rocket is launched from a tower. The height of the rocket,y in feet, is related to the time after launch, x in seconds, by the given equation. Using this equation, find out the time at which the rocket will reach its max, to the nearest 100th of a second. y=-16x^2+171x+87

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DonovanMaster · Tutor for 5 years

Answer

### The rocket will reach its maximum height at approximately 5.34 seconds.

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## Step 1: Identify the equation<br />### The height \( y \) as a function of time \( x \) is given by \( y = -16x^2 + 171x + 87 \).<br /><br />## Step 2: Find the time to reach the maximum height<br />### To find the time at which the rocket reaches its maximum height, use the formula for the vertex of a parabola given by \( ax^2 + bx + c \). The time \( x \) at the vertex is \( x = -\frac{b}{2a} \), where \( a = -16 \) and \( b = 171 \).<br /><br />## Step 3: Calculate the time<br />### Substitute \( a \) and \( b \) into the vertex formula:<br />\[ x = -\frac{171}{2 \times -16} = \frac{171}{32} \approx 5.34 \]
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