A soccer player kicks the ball to a height of 1 meter inside the goal. The equation for the height, h, of the ball at time, t, is h=-4.9t^2-5t+2. Find the time the ball reached the goal.

Respuesta :

1=-4.9t^2-5t+2
0=-4.9t^2-5t+1

5+square root of 25+39.2
Divided by -9.8
25+39.2=64.2

5+8 or 5-8
Dived by -9.8

-3/-9.8
0.3 seconds

By definition of the zeros of a quadratic function, the time the ball reached the goal is 0.1712.

Zeros of a function

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.

In a quadratic function that has the form:

f(x)= ax² + bx + c

the zeros or roots are calculated by:

[tex]x1,x2=\frac{-b+-\sqrt{b^{2}-4ac } }{2a}[/tex]

Time the ball reached the goal

The equation for the height h of the ball at time t is h=-4.9t²-5t+2.

A soccer player kicks the ball to a height of 1 meter inside the goal. This is, h=1. Replacing:

1=-4.9t²-5t+2

Solving:

0=-4.9t²-5t+2 -1

0=-4.9t²-5t+1

So, being:

  • a= -4.9
  • b= -5
  • c= 1

the zeros or roots are calculated as:

[tex]t1=\frac{-(-5)+\sqrt{(-5)^{2}-4x(-4.9)x1 } }{2x(-4.9)}[/tex]

[tex]t1=\frac{5+\sqrt{25 + 19.6 } }{-9.8}[/tex]

[tex]t1=\frac{5+\sqrt{44.6 } }{-9.8}[/tex]

[tex]t1=\frac{5+6.678 }{-9.8}[/tex]

[tex]t1=\frac{11.678 }{-9.8}[/tex]

t1= -1.192

and

[tex]t2=\frac{-(-5)-\sqrt{(-5)^{2}-4x(-4.9)x1 } }{2x(-4.9)}[/tex]

[tex]t2=\frac{5-\sqrt{25 + 19.6 } }{-9.8}[/tex]

[tex]t2=\frac{5-\sqrt{44.6 } }{-9.8}[/tex]

[tex]t2=\frac{5-6.678 }{-9.8}[/tex]

[tex]t2=\frac{-1.678}{-9.8}[/tex]

t2= 0.1712

Since time cannot be negative, the time the ball reached the goal is 0.1712.

Learn more about the zeros of a quadratic function:

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