Solved on Feb 27, 2024

An arrow is shot upward at 80 ft/s from a 25 ft platform. The path is $h=-16t^{2}+80t+25$ , where $h$ is height and $t$ is time. Find the maximum height.

STEP 1

Assumptions
1. The initial speed of the arrow is 80 feet per second.
2. The height of the platform from which the arrow is shot is 25 feet.
3. The pathway of the arrow is given by the equation $h = -16t^2 + 80t + 25$ , where $h$ is the height in feet and $t$ is the time in seconds.
4. The acceleration due to gravity is taken as $32$ feet per second squared in the downward direction, hence the $-16$ in the equation as $-16$ is half of $32$ .
5. We need to find the maximum height, which corresponds to the vertex of the parabola represented by the equation.

STEP 2

The equation $h = -16t^2 + 80t + 25$ is a quadratic equation in the form $h = at^2 + bt + c$ , which represents a parabola. The maximum height of the arrow corresponds to the vertex of the parabola.

STEP 3

The time at which the maximum height is reached can be found by using the formula for the vertex of a parabola given by $t = -\frac{b}{2a}$ .

STEP 4

Plug in the values for $a$ and $b$ from the equation $h = -16t^2 + 80t + 25$ into the vertex formula.
$a = -16, \quad b = 80$
$t = -\frac{b}{2a} = -\frac{80}{2(-16)}$

STEP 5

Calculate the time at which the maximum height is reached.
$t = -\frac{80}{-32} = \frac{80}{32} = 2.5 \text{ seconds}$

STEP 6

Now that we have the time at which the maximum height is reached, we can find the maximum height by plugging this time back into the original equation.

STEP 7

Plug $t = 2.5$ into the equation $h = -16t^2 + 80t + 25$ .
$h = -16(2.5)^2 + 80(2.5) + 25$

STEP 8

Calculate the value of $h$ to find the maximum height.
$h = -16(6.25) + 200 + 25$

STEP 9

Simplify the calculation.
$h = -100 + 200 + 25$

STEP 10

Calculate the maximum height.
$h = 100 + 25 = 125 \text{ feet}$
The maximum height of the arrow is 125 feet.

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An arrow is shot upward at 80 ft/s from a 25 ft platform. The path is $h=-16t^{2}+80t+25$ , where $h$ is height and $t$ is time. Find the maximum height.

STEP 1

STEP 2

The equation $h = -16t^2 + 80t + 25$ is a quadratic equation in the form $h = at^2 + bt + c$ , which represents a parabola. The maximum height of the arrow corresponds to the vertex of the parabola.

STEP 3

The time at which the maximum height is reached can be found by using the formula for the vertex of a parabola given by $t = -\frac{b}{2a}$ .

STEP 4

Plug in the values for $a$ and $b$ from the equation $h = -16t^2 + 80t + 25$ into the vertex formula.
$a = -16, \quad b = 80$
$t = -\frac{b}{2a} = -\frac{80}{2(-16)}$

STEP 5

Calculate the time at which the maximum height is reached.
$t = -\frac{80}{-32} = \frac{80}{32} = 2.5 \text{ seconds}$

STEP 6

Now that we have the time at which the maximum height is reached, we can find the maximum height by plugging this time back into the original equation.

STEP 7

Plug $t = 2.5$ into the equation $h = -16t^2 + 80t + 25$ .
$h = -16(2.5)^2 + 80(2.5) + 25$

STEP 8

Calculate the value of $h$ to find the maximum height.
$h = -16(6.25) + 200 + 25$

STEP 9

Simplify the calculation.
$h = -100 + 200 + 25$

STEP 10

Calculate the maximum height.
$h = 100 + 25 = 125 \text{ feet}$
The maximum height of the arrow is 125 feet.

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An arrow is shot upward at 80 ft/s from a 25 ft platform. The path is h=−16t2+80t+25h=-16t^{2}+80t+25h=−16t2+80t+25, where hhh is height and ttt is time. Find the maximum height.

STEP 1

STEP 2

The equation h=−16t2+80t+25 h = -16t^2 + 80t + 25 h=−16t2+80t+25 is a quadratic equation in the form h=at2+bt+c h = at^2 + bt + c h=at2+bt+c, which represents a parabola. The maximum height of the arrow corresponds to the vertex of the parabola.

STEP 3

The time at which the maximum height is reached can be found by using the formula for the vertex of a parabola given by t=−b2a t = -\frac{b}{2a} t=−2ab​.

STEP 4

Plug in the values for a a a and b b b from the equation h=−16t2+80t+25 h = -16t^2 + 80t + 25 h=−16t2+80t+25 into the vertex formula.a=−16,b=80 a = -16, \quad b = 80 a=−16,b=80t=−b2a=−802(−16) t = -\frac{b}{2a} = -\frac{80}{2(-16)} t=−2ab​=−2(−16)80​

STEP 5

Calculate the time at which the maximum height is reached.t=−80−32=8032=2.5 seconds t = -\frac{80}{-32} = \frac{80}{32} = 2.5 \text{ seconds} t=−−3280​=3280​=2.5 seconds

STEP 6

Now that we have the time at which the maximum height is reached, we can find the maximum height by plugging this time back into the original equation.

STEP 7

Plug t=2.5 t = 2.5 t=2.5 into the equation h=−16t2+80t+25 h = -16t^2 + 80t + 25 h=−16t2+80t+25.h=−16(2.5)2+80(2.5)+25 h = -16(2.5)^2 + 80(2.5) + 25 h=−16(2.5)2+80(2.5)+25

STEP 8

Calculate the value of h h h to find the maximum height.h=−16(6.25)+200+25 h = -16(6.25) + 200 + 25 h=−16(6.25)+200+25

STEP 9

Simplify the calculation.h=−100+200+25 h = -100 + 200 + 25 h=−100+200+25

STEP 10

Calculate the maximum height.h=100+25=125 feet h = 100 + 25 = 125 \text{ feet} h=100+25=125 feetThe maximum height of the arrow is 125 feet.

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An arrow is shot upward at 80 ft/s from a 25 ft platform. The path is $h=-16t^{2}+80t+25$ , where $h$ is height and $t$ is time. Find the maximum height.

The equation $h = -16t^2 + 80t + 25$ is a quadratic equation in the form $h = at^2 + bt + c$ , which represents a parabola. The maximum height of the arrow corresponds to the vertex of the parabola.

The time at which the maximum height is reached can be found by using the formula for the vertex of a parabola given by $t = -\frac{b}{2a}$ .

Plug in the values for $a$ and $b$ from the equation $h = -16t^2 + 80t + 25$ into the vertex formula.
$a = -16, \quad b = 80$
$t = -\frac{b}{2a} = -\frac{80}{2(-16)}$

Calculate the time at which the maximum height is reached.
$t = -\frac{80}{-32} = \frac{80}{32} = 2.5 \text{ seconds}$

Plug $t = 2.5$ into the equation $h = -16t^2 + 80t + 25$ .
$h = -16(2.5)^2 + 80(2.5) + 25$

Calculate the value of $h$ to find the maximum height.
$h = -16(6.25) + 200 + 25$

Simplify the calculation.
$h = -100 + 200 + 25$

Calculate the maximum height.
$h = 100 + 25 = 125 \text{ feet}$
The maximum height of the arrow is 125 feet.