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An object is thrown upward at a speed of 58 feet per second by a machine from a height of 7 feet off the ground. The height h of the object after t seconds can be found using the equation h=−16t^2+58t+7

a.When will the height be 17 feet? ______


b. When will the object reach the ground? ______

Respuesta :

Abinatamang

Answer:

First part:

Set h(t) = 17and solve for t.

-16t²+ 58t + 7= 17

-16t² + 58t - 10 = 0

Solve this quadratic equation for t. You should get 2 positive solutions. The lower value is the time to reach 17 on the way up, and the higher value is the time to reach 17 again, on the way down.

Second part:

Set h(t) = 0 and solve the resulting quadratic equation for t. You should get a negative solution (which you can discard), and a positive solution. The latter is your answer.

Ver imagen Abinatamang
boffeemadrid

The time required to reach 17 feet and the ground by the ball is required.

The time taken to reach 17 feet is 0.181 s.

To reach the ground the time taken is 3.74 s.

The equation is

[tex]h=-16t^2+58t+7[/tex]

[tex]h=17[/tex]

[tex]17=-16t^2+58t+7\\\Rightarrow -16t^2+58t-10=0\\\Rightarrow t=\frac{-58\pm \sqrt{58^2-4\left(-16\right)\left(-10\right)}}{2\left(-16\right)}\\\Rightarrow t=0.181,3.44[/tex]

The time taken reach a height of 17 feet while going up is 0.181 s.

On the ground [tex]h=0[/tex]

[tex]0=-16t^2+58t+7\\\Rightarrow t=\frac{-58\pm \sqrt{58^2-4\left(-16\right)\times 7}}{2\left(-16\right)}\\\Rightarrow t=-0.12,3.74[/tex]

The time taken to reach the ground is 3.74 s.

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