Math / Algebra

QuestionA ball is thrown from a height of 79 meters with an initial downward velocity of $6 \mathrm{~m} / \mathrm{s}$ . The ball's height $h$ (in meters) after $t$ seconds is given by the $f$ $h=79-6 t-5 t^{2}$
How long after the ball is thrown does it hit the ground? Round your answer(s) to the nearest hundredth. (If there is more than one answer, use the "or" button.) $t=$ Џ seconds

Studdy Solution

STEP 1

1. The height of the ball is given by the quadratic equation $h = 79 - 6t - 5t^2$ .
2. The ball hits the ground when its height $h$ is $0$ .
3. We need to solve the quadratic equation for $t$ when $h = 0$ .

STEP 2

1. Set the height equation equal to zero.
2. Rearrange the equation into standard quadratic form.
3. Use the quadratic formula to solve for $t$ .
4. Round the solution(s) to the nearest hundredth.
5. Interpret the solutions to find the time when the ball hits the ground.

STEP 3

Set the height equation equal to zero to find when the ball hits the ground:
$79 - 6t - 5t^2 = 0$

STEP 4

Rearrange the equation into standard quadratic form $at^2 + bt + c = 0$ :
$-5t^2 - 6t + 79 = 0$

STEP 5

Identify the coefficients for the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ :
- $a = -5$ - $b = -6$ - $c = 79$

STEP 6

Calculate the discriminant $b^2 - 4ac$ :
$(-6)^2 - 4(-5)(79) = 36 + 1580 = 1616$

STEP 7

Use the quadratic formula to solve for $t$ :
$t = \frac{-(-6) \pm \sqrt{1616}}{2(-5)}$ $t = \frac{6 \pm \sqrt{1616}}{-10}$

STEP 8

Calculate the square root and solve for $t$ :
$\sqrt{1616} \approx 40.20$
$t = \frac{6 \pm 40.20}{-10}$

STEP 9

Find the two possible values for $t$ :
$t_1 = \frac{6 + 40.20}{-10} = \frac{46.20}{-10} = -4.62$
$t_2 = \frac{6 - 40.20}{-10} = \frac{-34.20}{-10} = 3.42$

STEP 10

Round the positive solution to the nearest hundredth:
$t = 3.42$

STEP 11

Interpret the solution: Since time cannot be negative, the ball hits the ground at $t = 3.42$ seconds.
The ball hits the ground after:
$t = 3.42$ seconds

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Studdy Solution

STEP 1

1. The height of the ball is given by the quadratic equation $h = 79 - 6t - 5t^2$ .
2. The ball hits the ground when its height $h$ is $0$ .
3. We need to solve the quadratic equation for $t$ when $h = 0$ .

STEP 2

1. Set the height equation equal to zero.
2. Rearrange the equation into standard quadratic form.
3. Use the quadratic formula to solve for $t$ .
4. Round the solution(s) to the nearest hundredth.
5. Interpret the solutions to find the time when the ball hits the ground.

STEP 3

Set the height equation equal to zero to find when the ball hits the ground:
$79 - 6t - 5t^2 = 0$

STEP 4

Rearrange the equation into standard quadratic form $at^2 + bt + c = 0$ :
$-5t^2 - 6t + 79 = 0$

STEP 5

Identify the coefficients for the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ :
- $a = -5$ - $b = -6$ - $c = 79$

STEP 6

Calculate the discriminant $b^2 - 4ac$ :
$(-6)^2 - 4(-5)(79) = 36 + 1580 = 1616$

STEP 7

Use the quadratic formula to solve for $t$ :
$t = \frac{-(-6) \pm \sqrt{1616}}{2(-5)}$ $t = \frac{6 \pm \sqrt{1616}}{-10}$

STEP 8

Calculate the square root and solve for $t$ :
$\sqrt{1616} \approx 40.20$
$t = \frac{6 \pm 40.20}{-10}$

STEP 9

Find the two possible values for $t$ :
$t_1 = \frac{6 + 40.20}{-10} = \frac{46.20}{-10} = -4.62$
$t_2 = \frac{6 - 40.20}{-10} = \frac{-34.20}{-10} = 3.42$

STEP 10

Round the positive solution to the nearest hundredth:
$t = 3.42$

STEP 11

Interpret the solution: Since time cannot be negative, the ball hits the ground at $t = 3.42$ seconds.
The ball hits the ground after:
$t = 3.42$ seconds

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Studdy Solution

STEP 1

1. The height of the ball is given by the quadratic equation h=79−6t−5t2 h = 79 - 6t - 5t^2 h=79−6t−5t2.2. The ball hits the ground when its height h h h is 0 0 0.3. We need to solve the quadratic equation for t t t when h=0 h = 0 h=0.

STEP 2

1. Set the height equation equal to zero.2. Rearrange the equation into standard quadratic form.3. Use the quadratic formula to solve for t t t.4. Round the solution(s) to the nearest hundredth.5. Interpret the solutions to find the time when the ball hits the ground.

STEP 3

Set the height equation equal to zero to find when the ball hits the ground:79−6t−5t2=0 79 - 6t - 5t^2 = 0 79−6t−5t2=0

STEP 4

Rearrange the equation into standard quadratic form at2+bt+c=0 at^2 + bt + c = 0 at2+bt+c=0:−5t2−6t+79=0 -5t^2 - 6t + 79 = 0 −5t2−6t+79=0

STEP 5

Identify the coefficients for the quadratic formula t=−b±b2−4ac2a t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} t=2a−b±b2−4ac​​:- a=−5 a = -5 a=−5 - b=−6 b = -6 b=−6 - c=79 c = 79 c=79

STEP 6

Calculate the discriminant b2−4ac b^2 - 4ac b2−4ac:(−6)2−4(−5)(79)=36+1580=1616 (-6)^2 - 4(-5)(79) = 36 + 1580 = 1616 (−6)2−4(−5)(79)=36+1580=1616

STEP 7

Use the quadratic formula to solve for t t t:t=−(−6)±16162(−5) t = \frac{-(-6) \pm \sqrt{1616}}{2(-5)} t=2(−5)−(−6)±1616​​ t=6±1616−10 t = \frac{6 \pm \sqrt{1616}}{-10} t=−106±1616​​

STEP 8

Calculate the square root and solve for t t t:1616≈40.20 \sqrt{1616} \approx 40.20 1616​≈40.20t=6±40.20−10 t = \frac{6 \pm 40.20}{-10} t=−106±40.20​

STEP 9

STEP 10

Round the positive solution to the nearest hundredth:t=3.42 t = 3.42 t=3.42

STEP 11

Interpret the solution: Since time cannot be negative, the ball hits the ground at t=3.42 t = 3.42 t=3.42 seconds. The ball hits the ground after:t=3.42 t = 3.42 t=3.42 seconds

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1. The height of the ball is given by the quadratic equation $h = 79 - 6t - 5t^2$ .
2. The ball hits the ground when its height $h$ is $0$ .
3. We need to solve the quadratic equation for $t$ when $h = 0$ .

1. Set the height equation equal to zero.
2. Rearrange the equation into standard quadratic form.
3. Use the quadratic formula to solve for $t$ .
4. Round the solution(s) to the nearest hundredth.
5. Interpret the solutions to find the time when the ball hits the ground.

Set the height equation equal to zero to find when the ball hits the ground:
$79 - 6t - 5t^2 = 0$

Rearrange the equation into standard quadratic form $at^2 + bt + c = 0$ :
$-5t^2 - 6t + 79 = 0$

Identify the coefficients for the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ :
- $a = -5$ - $b = -6$ - $c = 79$

Calculate the discriminant $b^2 - 4ac$ :
$(-6)^2 - 4(-5)(79) = 36 + 1580 = 1616$

Use the quadratic formula to solve for $t$ :
$t = \frac{-(-6) \pm \sqrt{1616}}{2(-5)}$ $t = \frac{6 \pm \sqrt{1616}}{-10}$

Calculate the square root and solve for $t$ :
$\sqrt{1616} \approx 40.20$
$t = \frac{6 \pm 40.20}{-10}$

Round the positive solution to the nearest hundredth:
$t = 3.42$

Interpret the solution: Since time cannot be negative, the ball hits the ground at $t = 3.42$ seconds.
The ball hits the ground after:
$t = 3.42$ seconds