Math / Algebra

QuestionNASA launches a rocket at $t=0$ seconds. Its height, in meters above sea-level, as a function of time is given by $h(t)=-4.9 t^{2}+40 t+185$ .
Assuming that the rocket will splash down into the ocean, at that time does splashdown occur? The rocket splashes down after $\square$ seconds. (Round your answer to 2 decimals.)
How high above sea-level does the rocket get at its peak? The rocket peaks at $\square$ 266.63 0 meters above sea-level. (Round your answer to 2 decimals.) Question Help: $\square$ Video 1 $\square$ Video 2 $\square$ Message instructor Submit Question

Studdy Solution

STEP 1

1. The rocket's height as a function of time is given by a quadratic equation: $h(t) = -4.9t^2 + 40t + 185$ .
2. The rocket splashes down when its height $h(t)$ is zero.
3. The peak height occurs at the vertex of the parabola described by the quadratic function.

STEP 2

1. Determine the time of splashdown by finding when $h(t) = 0$ .
2. Determine the time at which the rocket reaches its peak height.
3. Calculate the peak height of the rocket.

STEP 3

Set the height function equal to zero to find the time of splashdown:
$-4.9t^2 + 40t + 185 = 0$

STEP 4

Use the quadratic formula to solve for $t$ . The quadratic formula is given by:
$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
where $a = -4.9$ , $b = 40$ , and $c = 185$ .

STEP 5

Calculate the discriminant $b^2 - 4ac$ :
$b^2 - 4ac = 40^2 - 4(-4.9)(185)$ $= 1600 + 3626$ $= 5226$

STEP 6

Substitute the values into the quadratic formula:
$t = \frac{-40 \pm \sqrt{5226}}{2(-4.9)}$

STEP 7

Calculate the two possible values for $t$ :
$t = \frac{-40 + \sqrt{5226}}{-9.8} \quad \text{and} \quad t = \frac{-40 - \sqrt{5226}}{-9.8}$

STEP 8

Since time cannot be negative, we only consider the positive value:
$t \approx \frac{-40 + 72.31}{-9.8}$ $t \approx \frac{32.31}{9.8}$ $t \approx 3.30$

STEP 9

To find the time at which the rocket reaches its peak height, use the vertex formula for a parabola $t = -\frac{b}{2a}$ :
$t = -\frac{40}{2(-4.9)}$ $t = \frac{40}{9.8}$ $t \approx 4.08$

STEP 10

Substitute $t = 4.08$ back into the height function to find the peak height:
$h(4.08) = -4.9(4.08)^2 + 40(4.08) + 185$

STEP 11

Calculate the peak height:
$h(4.08) \approx -4.9(16.6464) + 163.2 + 185$ $h(4.08) \approx -81.57 + 163.2 + 185$ $h(4.08) \approx 266.63$
The rocket splashes down after $\boxed{3.30}$ seconds. The rocket peaks at $\boxed{266.63}$ meters above sea-level.

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Studdy Solution

STEP 1

1. The rocket's height as a function of time is given by a quadratic equation: $h(t) = -4.9t^2 + 40t + 185$ .
2. The rocket splashes down when its height $h(t)$ is zero.
3. The peak height occurs at the vertex of the parabola described by the quadratic function.

STEP 2

1. Determine the time of splashdown by finding when $h(t) = 0$ .
2. Determine the time at which the rocket reaches its peak height.
3. Calculate the peak height of the rocket.

STEP 3

Set the height function equal to zero to find the time of splashdown:
$-4.9t^2 + 40t + 185 = 0$

STEP 4

Use the quadratic formula to solve for $t$ . The quadratic formula is given by:
$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
where $a = -4.9$ , $b = 40$ , and $c = 185$ .

STEP 5

Calculate the discriminant $b^2 - 4ac$ :
$b^2 - 4ac = 40^2 - 4(-4.9)(185)$ $= 1600 + 3626$ $= 5226$

STEP 6

Substitute the values into the quadratic formula:
$t = \frac{-40 \pm \sqrt{5226}}{2(-4.9)}$

STEP 7

Calculate the two possible values for $t$ :
$t = \frac{-40 + \sqrt{5226}}{-9.8} \quad \text{and} \quad t = \frac{-40 - \sqrt{5226}}{-9.8}$

STEP 8

Since time cannot be negative, we only consider the positive value:
$t \approx \frac{-40 + 72.31}{-9.8}$ $t \approx \frac{32.31}{9.8}$ $t \approx 3.30$

STEP 9

To find the time at which the rocket reaches its peak height, use the vertex formula for a parabola $t = -\frac{b}{2a}$ :
$t = -\frac{40}{2(-4.9)}$ $t = \frac{40}{9.8}$ $t \approx 4.08$

STEP 10

Substitute $t = 4.08$ back into the height function to find the peak height:
$h(4.08) = -4.9(4.08)^2 + 40(4.08) + 185$

STEP 11

Calculate the peak height:
$h(4.08) \approx -4.9(16.6464) + 163.2 + 185$ $h(4.08) \approx -81.57 + 163.2 + 185$ $h(4.08) \approx 266.63$
The rocket splashes down after $\boxed{3.30}$ seconds. The rocket peaks at $\boxed{266.63}$ meters above sea-level.

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Studdy Solution

STEP 1

STEP 2

1. Determine the time of splashdown by finding when h(t)=0 h(t) = 0 h(t)=0.2. Determine the time at which the rocket reaches its peak height.3. Calculate the peak height of the rocket.

STEP 3

Set the height function equal to zero to find the time of splashdown:−4.9t2+40t+185=0 -4.9t^2 + 40t + 185 = 0 −4.9t2+40t+185=0

STEP 4

Use the quadratic formula to solve for t t t. The quadratic formula is given by:t=−b±b2−4ac2a t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} t=2a−b±b2−4ac​​where a=−4.9 a = -4.9 a=−4.9, b=40 b = 40 b=40, and c=185 c = 185 c=185.

STEP 5

Calculate the discriminant b2−4ac b^2 - 4ac b2−4ac:b2−4ac=402−4(−4.9)(185) b^2 - 4ac = 40^2 - 4(-4.9)(185) b2−4ac=402−4(−4.9)(185) =1600+3626 = 1600 + 3626 =1600+3626 =5226 = 5226 =5226

STEP 6

Substitute the values into the quadratic formula:t=−40±52262(−4.9) t = \frac{-40 \pm \sqrt{5226}}{2(-4.9)} t=2(−4.9)−40±5226​​

STEP 7

Calculate the two possible values for t t t:t=−40+5226−9.8andt=−40−5226−9.8 t = \frac{-40 + \sqrt{5226}}{-9.8} \quad \text{and} \quad t = \frac{-40 - \sqrt{5226}}{-9.8} t=−9.8−40+5226​​andt=−9.8−40−5226​​

STEP 8

Since time cannot be negative, we only consider the positive value:t≈−40+72.31−9.8 t \approx \frac{-40 + 72.31}{-9.8} t≈−9.8−40+72.31​ t≈32.319.8 t \approx \frac{32.31}{9.8} t≈9.832.31​ t≈3.30 t \approx 3.30 t≈3.30

STEP 9

To find the time at which the rocket reaches its peak height, use the vertex formula for a parabola t=−b2a t = -\frac{b}{2a} t=−2ab​:t=−402(−4.9) t = -\frac{40}{2(-4.9)} t=−2(−4.9)40​ t=409.8 t = \frac{40}{9.8} t=9.840​ t≈4.08 t \approx 4.08 t≈4.08

STEP 10

Substitute t=4.08 t = 4.08 t=4.08 back into the height function to find the peak height:h(4.08)=−4.9(4.08)2+40(4.08)+185 h(4.08) = -4.9(4.08)^2 + 40(4.08) + 185 h(4.08)=−4.9(4.08)2+40(4.08)+185

STEP 11

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1. Determine the time of splashdown by finding when $h(t) = 0$ .
2. Determine the time at which the rocket reaches its peak height.
3. Calculate the peak height of the rocket.

Set the height function equal to zero to find the time of splashdown:
$-4.9t^2 + 40t + 185 = 0$

Use the quadratic formula to solve for $t$ . The quadratic formula is given by:
$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
where $a = -4.9$ , $b = 40$ , and $c = 185$ .

Calculate the discriminant $b^2 - 4ac$ :
$b^2 - 4ac = 40^2 - 4(-4.9)(185)$ $= 1600 + 3626$ $= 5226$

Substitute the values into the quadratic formula:
$t = \frac{-40 \pm \sqrt{5226}}{2(-4.9)}$

Calculate the two possible values for $t$ :
$t = \frac{-40 + \sqrt{5226}}{-9.8} \quad \text{and} \quad t = \frac{-40 - \sqrt{5226}}{-9.8}$

Since time cannot be negative, we only consider the positive value:
$t \approx \frac{-40 + 72.31}{-9.8}$ $t \approx \frac{32.31}{9.8}$ $t \approx 3.30$

To find the time at which the rocket reaches its peak height, use the vertex formula for a parabola $t = -\frac{b}{2a}$ :
$t = -\frac{40}{2(-4.9)}$ $t = \frac{40}{9.8}$ $t \approx 4.08$

Substitute $t = 4.08$ back into the height function to find the peak height:
$h(4.08) = -4.9(4.08)^2 + 40(4.08) + 185$