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Answer :
To solve the problem of finding the maximum height reached by the toy rocket, we start with the given height equation:
[tex]\[ h(x) = -16x^2 + 128x \][/tex]
where [tex]\( h(x) \)[/tex] is the height of the rocket in feet and [tex]\( x \)[/tex] is the time in seconds.
### Step 1: Find the time at which the rocket reaches its maximum height
The maximum height occurs when the vertical velocity becomes zero, which is the peak of the projectile's trajectory. This means we need to find the value of [tex]\( x \)[/tex] for which the derivative of [tex]\( h(x) \)[/tex] with respect to [tex]\( x \)[/tex] equals zero.
1. Differentiate [tex]\( h(x) \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ h'(x) = \frac{d}{dx}(-16x^2 + 128x) = -32x + 128 \][/tex]
2. Set [tex]\( h'(x) \)[/tex] equal to zero to find the critical points:
[tex]\[ -32x + 128 = 0 \][/tex]
3. Solve for [tex]\( x \)[/tex]:
[tex]\[ -32x = -128 \][/tex]
[tex]\[ x = \frac{128}{32} \][/tex]
[tex]\[ x = 4 \][/tex]
Thus, the rocket reaches its maximum height at [tex]\( x = 4 \)[/tex] seconds.
### Step 2: Calculate the maximum height
Now that we have the time at which the maximum height is reached, substitute [tex]\( x = 4 \)[/tex] back into the original height equation to find the maximum height.
[tex]\[ h(4) = -16(4)^2 + 128(4) \][/tex]
1. Calculate [tex]\( 4^2 \)[/tex]:
[tex]\[ 4^2 = 16 \][/tex]
2. Substitute back:
[tex]\[ h(4) = -16 \times 16 + 128 \times 4 \][/tex]
[tex]\[ h(4) = -256 + 512 \][/tex]
[tex]\[ h(4) = 256 \][/tex]
Therefore, the maximum height reached by the toy rocket is 256 feet.
In conclusion, the rocket reaches its maximum height of 256 feet at 4 seconds after it is launched.
[tex]\[ h(x) = -16x^2 + 128x \][/tex]
where [tex]\( h(x) \)[/tex] is the height of the rocket in feet and [tex]\( x \)[/tex] is the time in seconds.
### Step 1: Find the time at which the rocket reaches its maximum height
The maximum height occurs when the vertical velocity becomes zero, which is the peak of the projectile's trajectory. This means we need to find the value of [tex]\( x \)[/tex] for which the derivative of [tex]\( h(x) \)[/tex] with respect to [tex]\( x \)[/tex] equals zero.
1. Differentiate [tex]\( h(x) \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ h'(x) = \frac{d}{dx}(-16x^2 + 128x) = -32x + 128 \][/tex]
2. Set [tex]\( h'(x) \)[/tex] equal to zero to find the critical points:
[tex]\[ -32x + 128 = 0 \][/tex]
3. Solve for [tex]\( x \)[/tex]:
[tex]\[ -32x = -128 \][/tex]
[tex]\[ x = \frac{128}{32} \][/tex]
[tex]\[ x = 4 \][/tex]
Thus, the rocket reaches its maximum height at [tex]\( x = 4 \)[/tex] seconds.
### Step 2: Calculate the maximum height
Now that we have the time at which the maximum height is reached, substitute [tex]\( x = 4 \)[/tex] back into the original height equation to find the maximum height.
[tex]\[ h(4) = -16(4)^2 + 128(4) \][/tex]
1. Calculate [tex]\( 4^2 \)[/tex]:
[tex]\[ 4^2 = 16 \][/tex]
2. Substitute back:
[tex]\[ h(4) = -16 \times 16 + 128 \times 4 \][/tex]
[tex]\[ h(4) = -256 + 512 \][/tex]
[tex]\[ h(4) = 256 \][/tex]
Therefore, the maximum height reached by the toy rocket is 256 feet.
In conclusion, the rocket reaches its maximum height of 256 feet at 4 seconds after it is launched.
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