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Please Help Me !!!! Its A Math Question!! The path of a ball being thrown off a building can be modeled by the equation y= -16t^2+64t+80

How long does it take to get to its maximum height?

What is its maximum height?

How long does it take to hit the ground?

What is an appropriate domain and range for this function?

What does the 80 in the equation represent?

Please Help Me Its A Math Question The path of a ball being thrown off a building can be modeled by the equation y 16t264t80 How long does it take to get to its class=

Respuesta :

Answer: Also can i have brainliest?

The ball will hit the ground 5 seconds after being thrown.

Step-by-step 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= t^2-4t-5

To find the solution of the Quadratic equation, you can factor 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)

t1=5

t2=-1

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

For the path of a ball being thrown maximum height is 144 units, time taken to reach maximum height is 2 units and time, it take to hit the ground is 5 units.

What is projectile motion?

Projectile motion is the motion of the body, when it is thrown in the air taking the action of gravity on it.

The path of a ball being thrown off a building can be modeled by the equation

[tex]y= -16t^2+64t+80[/tex]

  • Time taken to reach maximum height-

Comparing this equation, with standard form of quadratic equation, we get,  a=-16, b=64 and c=80.

Thus, the maximum height time can be given as,

[tex]t=-\dfrac{b}{2a}\\t=-\dfrac{64}{2(-16)}\\t=2\rm sec[/tex]

Thus, the time taken to reach maximum height is 2 seconds.

  • The maximum height-

Again by the equation,

[tex]y= -16t^2+64t+80[/tex]
As it reaches maximum height after the 2 seconds. Therefore,

[tex]y_{max}= -16(2)^2+64(2)+80\\y_{max}= 144\rm[/tex]

  • Time, it take to hit the ground-

This can be find out with the following formula,

[tex]t=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\t=\dfrac{-64\pm\sqrt{64^2-4(-16)(80)}}{2(-16)}\\t=\dfrac{-64\pm96}{-32}\\[/tex]

Taking negative sign,

[tex]t=\dfrac{-64-96}{-32}=5\\[/tex]

Taking positive sign,

[tex]t=\dfrac{-64+96}{-32}=-1\\[/tex]

Thus the time, it take to hit the ground is 5 units.

  • The 80 in the equation represent-

80 is the constant coefficient of the equation, which represent the height in the feet.

Hence, for the path of a ball being thrown maximum height is 144 units, time taken to reach maximum height is 2 units and time, it take to hit the ground is 5 units.

Learn more about the projectile motion here;

https://brainly.com/question/24216590

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