The time taken by the rock to reach the lake below is 5.123 sec
t= 5.123 sec
What is quadratic equation?
A quadratic equation is an algebraic expression of the second degree in x. The quadratic equation in its standard form is [tex]ax^{2} +bx+c=0[/tex] where a, b are the coefficients, x is the variable, and c is the constant term.
It is given that,
H= [tex]-16t^{2} +32t+256[/tex], which represent the height of the rock
Also, we have given that the height of the cliff is 256 feet.
Now to find the time taken by the rock, take H=0
[tex]-16t^{2} +32t+256[/tex]=0
[tex]-16(t^{2} -2t-16) = 0[/tex]
[tex]t^{2} -2t-16 = 0[/tex]
- Solve the above quadratic equation using Discriminant Method
[tex]t= \frac{-(b)\pm\sqrt{(b)^{2}- 4 *a*c } )}{2*a}[/tex]
From above equation, a=1, b=-2, c=-16
So,
[tex]t= \frac{-(-2)\pm\sqrt{(-2)^{2}- 4 *1*(-16) } )}{2*1}[/tex]
[tex]t= \frac{2\pm\sqrt{(4+64 } )}{2}[/tex]
[tex]t= \frac{2\pm\sqrt{(68 } )}{2}[/tex]
There will be two value one with positive [tex]t= -2+\sqrt{(68 } )[/tex] and another with negative [tex]t= -2-\sqrt{(68 } )[/tex]
We will neglect negative value because time can't be negative.
[tex]t= \frac{2+\sqrt{(68 } )}{2}[/tex]
[tex]t= \frac{2+8.246}{2}[/tex]
[tex]t= \frac{10.246}{2}[/tex]
t= 5.123 sec
The time taken by the rock to reach the lake below is t= 5.123 sec
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