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AlekseyPX
3 years ago
15

3. A water balloon was thrown

Mathematics
1 answer:
Mazyrski [523]3 years ago
6 0

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.

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