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A ball is throw into the air from the top of a building. The height h(t) of the ball above the ground t seconds after it is thrown can be modeled by h (t) =-16t +64t+80 how many seconds after being thrown will the ball hit the ground

User Unikorn
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Answer: The ball will hit the ground 5 seconds after being thrown.

Explanation:

The correct function is:
h (t) =-16t^2 +64t+80

You can rewrite the Quadratic function given in the exercise with making
h(t)=0. Then this is:


0=-16t^2 +64t+80

Now you can simplify the equation dividing both sides by -16. So you get:


(0)/(-16)=(-16t^2 +64t+80)/(-16)\\\\0=t^2-4t-5

To find the solution of the Quadratic equation, you can facfor it. In order to do this, it is necessary to find two numbers whose sum is -4 and whose product is -5. These number would be -5 and 1.

Therefore, you get this result:


0=(t-5)(t+1)\\\\t_1=5\\t_2=-1

Since the time cannot be negative, you can conclude that the ball will hit the ground 5 seconds after being thrown.

User Yakob Abada
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7.9k points
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