Question

3. A water balloon was thrown

Answer 1

Note: It the given equation the coefficient of $x^2$ must be -16 instead of 16.

Given:

Consider the height  of the water balloon over  time can be modeled by the  function

$y=-16x^2+160x+50$

To find:

The maximum height of the water balloon  after it was thrown.

Solution:

We have,

$y=-16x^2+160x+50$

Here, leading coefficient is negative. So, it is a downward parabola and vertex of a downward parabola, is the point of maxima.

If a parabola is $f(x)=ax^2+bx+c$, then

$Vertex=\left(-\dfrac{b}{2a},f(-\dfrac{b}{2a})\right)$

Here, $a=-16,b=160,c=50$. So,

$-\dfrac{b}{2a}=-\dfrac{160}{2(-16)}$

$-\dfrac{b}{2a}=-\dfrac{160}{-32}$

$-\dfrac{b}{2a}=5$

Now, put x=5 in the given equation.

$y=-16(5)^2+160(5)+50$

$y=-16(25)+800+50$

$y=-400+850$

$y=450$

The vertex of the given parabolic equation is (5,450).

Therefore, the maximum height of the balloon is 450 units after 5 units of time.