Question

From a height of 256 feet above a lake on a cliff, Mikaela throws a rock out over the lake. The height of
the rock + seconds after Mikaela throws it is represented by the equation H=-16t2 + 32+ + 256. To the nearest tenth
of a second, how long does it take the rock to reach the lake below? (Hint: replace h with 0)

Answer 1

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 $ax^{2} +bx+c=0$ where a, b are the coefficients, x is the variable, and c is the constant term.

It is given that,

H= $-16t^{2} +32t+256$, 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

$-16t^{2} +32t+256$=0

$-16(t^{2} -2t-16) = 0$

$t^{2} -2t-16 = 0$

• Solve the above quadratic equation using Discriminant Method

$t= \frac{-(b)\pm\sqrt{(b)^{2}- 4 *a*c } )}{2*a}$

From above equation, a=1, b=-2, c=-16

So,

$t= \frac{-(-2)\pm\sqrt{(-2)^{2}- 4 *1*(-16) } )}{2*1}$

$t= \frac{2\pm\sqrt{(4+64 } )}{2}$

$t= \frac{2\pm\sqrt{(68 } )}{2}$

There will be two value one with positive $t= -2+\sqrt{(68 } )$ and another with negative $t= -2-\sqrt{(68 } )$

We will neglect negative value because time can't be negative.

$t= \frac{2+\sqrt{(68 } )}{2}$

$t= \frac{2+8.246}{2}$

$t= \frac{10.246}{2}$

t= 5.123 sec

The time taken by the rock to reach the lake below is t= 5.123 sec

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