Question

Keisha throws a rock down an old well the distance D in Feet The Rock Falls after T seconds can be represented by D =16t ^ 2 + 64t if the water in the well is 80 ft below ground how many seconds will it take for the rock to hit the water?​

Answer 1

Answer:

1 second

Step-by-step explanation:

16t^2+64t=80

16t^2+64t-80=0

64^2-4(16)(-80)

9216

-64-96

______ = -5

32

-64+96

_______ = 1 sec

32

Answer 2

In 1 second the rock to hit the water

What is an equation?

An equation represents any two expressions or statements are equal that means they are having an equal to (=) sign in between them.

What is a quadratic equation?

An equation whose highest degree is 2, is called a quadratic equation.

How to solve a quadratic equation?

We can solve a quadratic equation by two ways:

Factorization- If the quadratic expression is factorizable then we can factorize it and equate it to 0. Then we can apply the concept that when product of two or more expressions is 0 then at least one of the expression or both must be 0

By Shreedhar Acharya formula- A quadratic equation of the  form$ax^{2} +bx+c=0$  can be solved by the formula given below:

$x=\frac{-b + \sqrt{b^{2}-4ac } }{2a}$

or

$x=\frac{-b - \sqrt{b^{2}-4ac } }{2a}$

How to find how many seconds will it take for the rock to hit the water ?

According to the problem,  

•        Keisha throws a rock down an old well the distance D in Feet The Rock Falls after t seconds can be represented by $D=16t^{2} +64t$

• Here, D is 80 feet

We can write,  

$80=16t^{2} +64t$

⇒ $t^{2} + 4t - 5 = 0$

Solving by method of factorization,

⇒ $t^{2} - t + 5t - 5 =0$

⇒ (t-1)(t+5) =0

So, t=1 or t = -5

• Time cannot be negative

So, t = 1 sec

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