Quadratic applications question

In summary, the projectile will return to the ground after 20 seconds in the air, as determined by solving the equation -16t^2 + 320t = 0 algebraically. It is important to note that the value of t cannot be 0, as this would indicate the point of firing rather than the return to the ground.
  • #1
ghostchaox
4
0

Homework Statement


A projectile is launched upward. Its height above the ground after t seconds is -16t^2 + 320t. After how many seconds in the air will it return to the ground? Solve algebraically.


Homework Equations


ax^2+bx+c=0 ?
-b (+/-) (Square root of b^2 -4ac)/2a ?


The Attempt at a Solution



t= -320 (+/-) (Square root of 320^2 -4(-16)(0)

t= -320 (+/-) 320/ -16
t= {0,20} ?

If i check it, it doesn't fit.
 
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  • #2
Its simpler this way.

[itex]0=-16t^2+320t[/itex]

Divide both sides by -16

[itex]0=t^2-20t[/itex]

[itex]0=t(t-20)[/itex]

[tex]t=0,20[/tex]

We get the same solutions, check it the long way by paper, maybe your calculators got errors. My solutions are verified by my calculator..
 
  • #3
20=-16(20^2) + 320(20) ?
20= -6400+6400?
20=0?

20 =/= 0

0= -16(0^2) + 320(0) ?
0=0+0?
0=0


Now, since this is geometry, t(seconds) cannot equal 0 or else the time and distance would be also 0

If i plug in 20 as t, it ends up as 20=0, which is not true.

I don't understand how I am right
 
  • #4
t(seconds) cannot equal 0 or else the time and distance would be also 0

Why not? t=0 is just the point of firing - one of two points described by the parabola where the height is zero. What other value of t allows the equation:

-16t^2 + 320t = 0

to equal zero? This:

20=-16(20^2) + 320(20)

Is wrong. Why have you placed 20 on the left hand side?
 
  • #5
Right... lol Thanks!~!~
 

What is a quadratic application question?

A quadratic application question is a problem that involves using the quadratic formula or solving a quadratic equation to find the solution. These types of questions often involve real-world scenarios, such as calculating the height of an object or determining the maximum profit for a business.

How do I know when to use the quadratic formula?

You should use the quadratic formula when you have a quadratic equation in the form of ax^2 + bx + c = 0, where a, b, and c are constants. This formula can be used to solve for the values of x that make the equation true.

What are some common applications of quadratic equations?

Quadratic equations have many real-world applications, such as calculating the trajectory of a projectile, determining the maximum or minimum value of a function, and finding the dimensions of a rectangle with a given perimeter and area.

What are the steps for solving a quadratic equation?

The first step is to make sure the equation is in the form of ax^2 + bx + c = 0. Then, you can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a. Plug in the values for a, b, and c and simplify to find the solutions for x. Finally, check your answers by plugging them back into the original equation.

How can I apply quadratic equations to real-world problems?

To apply quadratic equations to real-world problems, you can start by identifying the known and unknown variables in the problem. Then, use the given information to set up a quadratic equation and solve using the quadratic formula. Finally, interpret the solutions in the context of the problem to find the answer.

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