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An object is launched from the ground. The object’s height, in feet, can be described by the quadratic function h(t) = 80t – 16t2, where t is the time, in seconds, since the object was launched. When will the object hit the ground after it is launched? Explain how you found your answer.

Respuesta :

corinaco

Sample Response: The object will hit the ground after 5 seconds. You can rewrite the quadratic function as a quadratic equation set equal to zero to find the zeros of the function 0 = –16t2 + 80t + 0. You can factor or use the quadratic formula to get t = 0 and t = 5. Therefore, it is on the ground at t = 0 (time of launch) and then hits the ground at t = 5 seconds

What did you include in your response? Check all that apply.

I Rewrite the quadratic function as a quadratic equation set equal to zero: 0 = –16t2 + 80t + 0.

Use the quadratic formula to solve for the zeros.

Factor to solve for the zeros.

t = 0 and t = 5 seconds.

The object will hit the ground after 5 seconds.

abidemiokin

This question deals with the application of intercepts in quadratic equations. It will take 5 seconds for the object to hit the ground.

Quadratic expressions

This question deals with the application of intercepts in quadratic equations. Given the function of the height reached by the object expressed as:

h(t) = 80t - 16t^2

The height of the object on the ground occurs at h(t) = 0. Substitute

80t - 16t^2 =0

16t^2= 80t

16t = 80

t = 80/16

t = 5secs

Hence it will take 5 seconds for the object to hit the ground.

Learn more on intercepts here: https://brainly.com/question/1884491

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