The height "h" of a ball thrown straight up with a velocity of 88 ft/s is given by h = -16t^2 + 88t where "t" is the time it is

Question

seropon [69]

2 years ago

7

The height "h" of a ball thrown straight up with a velocity of 88 ft/s is given by h = -16t^2 + 88t where "t" is the time it is

in the air. For how many seconds the ball is in the air before it hits the ground?

Mathematics

1 answer:

Dvinal [7] · Answer 1 · 2023-07-03T20:24:20+03:00

By definition of the zeros of ta quadratic function, for 5.5 seconds the ball is in the air before it hits the ground.

<h3>Zeros of a function</h3>

The points where a polynomial function crosses the axis of the independent term (x) represent the so-called zeros of the function.

That is, the zeros represent the roots of the polynomial equation that is obtained by making f(x)=0.

In summary, the roots or zeros of the quadratic function are those values of x for which the expression is equal to 0. Graphically, the roots correspond to the abscissa of the points where the parabola intersects the x-axis.

<h3>Time the ball is in the air before it hits the ground</h3>

In this case, the height "h" of a ball thrown straight up with a velocity of 88 ft/s is given by h = -16t² + 88t, where "t" is the time it is in the air.

When the ball hits he ground, he height h has a value of zero. This is h=0. Replacing in the previous expression for the height you get:

0= -16t² + 88t

It can be solved by extracting the term "t" as a common factor:

0= t×(-16t + 88)

The Zero Product Principle says that if the product of two numbers is 0, then at least one of the factors is 0. Then:

t= 0

or

0= -16t + 88

Solving: -88= -16t

(-88)÷ (-16)=t

Finally, this means that for 5.5 seconds the ball is in the air before it hits the ground.

Learn more about the zeros of a quadratic function: