Word Problems with Quadratic equations

[ohilveti9]

Hey guys I have a test coming up and i need help with these questions assigned for review



an object dropped from the top of a 45-metre tower falls according to the formula d = -5t^2+45. where *d* represent the distance of the object above the ground at a time, *t*, in seconds.

it doesnt even ask for anything but I assume how long will it take for it to reach the ground. the question says *d* represents the height at which the object is dropped? but how does that work as a quadratic equation? 45 = -5t^2+45. If you bring the 5t^2 over, you get get 5t^2 = 0, because the 45's cancel.

[Baltuilhe]

Good afternoon!

d is the distance above the ground, so:
[tex]d=-5t^2+45[/tex]

For t=0:
[tex]d=-5(0)^2+45=45[/tex]

The initial distance is d=45m

The end is when reaches the ground, or, d=0, so:
[tex]d=-5t^2+45[/tex]
[tex]0=-5t^2+45[/tex]
[tex]5t^2=45[/tex]
[tex]t^2=\frac{45}{5}=9[/tex]
[tex]t=3[/tex]

So, it reaches in 3s the ground!

[Guest]

Hey guys I have a test coming up and i need help with these questions assigned for review



an object dropped from the top of a 45-metre tower falls according to the formula d = -5t^2+45. where *d* represent the distance of the object above the ground at a time, *t*, in seconds.

it doesnt even ask for anything but I assume how long will it take for it to reach the ground. the question says *d* represents the height at which the object is dropped?

There is your mistake! The question does NOT say that! It says "*d* represents the distance of the object above the ground at a time, *t*, in seconds". The "height at which the object is dropped" is d(0) since it was dropped at time t= 0.

but how does that work as a quadratic equation? 45 = -5t^2+45. If you bring the 5t^2 over, you get get 5t^2 = 0, because the 45's cancel.

Yes, of course. d= 45 at the top of the tower when the object is first dropped, t= 0. When it "reaches the ground, d= 0. To find how long that takes you need to solve "-5t^2+ 45= 0", not 45.