Respuesta :
Answer:
t = 1s and t = 2s
Step-by-step explanation:
We have the equation
[tex]h(t)=vt - 16t^2[/tex]
where [tex]v=48ft/s[/tex]
and [tex]h=32ft[/tex]
Thus, the equation now becomes:
[tex]32=48t - 16t^2[/tex]
To solve, we must first equal the quadratic equation to zero
[tex]- 16t^2+48t -32=0[/tex]
Solve this equation for t using the general formula for a quadratic equation
[tex]x=\frac{-b+-\sqrt{ b^2-4ac} }{2a}[/tex]
but in this case intead of x we have t.
[tex]t=\frac{-b+-\sqrt{ b^2-4ac} }{2a}[/tex]
Where a is the coefficient of the quadratic term.
b is the coefficient of the linear term, and c the independent term.
[tex]a= -16\\b= 48\\c=-34[/tex]
And now we substitute these values in the general formula:
[tex]t=\frac{-48+-\sqrt{ 48^2-4(-16)(-32)} }{2(-16)}[/tex]
[tex]t=\frac{-48+-\sqrt{ 2304-2048} }{-32}[/tex]
[tex]t=\frac{-48+-\sqrt{256} }{-32}[/tex]
[tex]t=\frac{-48+-16 }{-32}[/tex]
Now, we are going to have two values for t due to the +- sign. Using the negative sign:
[tex]t=\frac{-48-16 }{-32}[/tex]
[tex]t=2s[/tex]
And using the positive sign:
[tex]t=\frac{-48+16 }{-32}[/tex]
[tex]t=1s[/tex]
At times t = 1s and t = 2s the object will it reach a height of 32 feet above the ground.
