Algebraic Approach to Calculating Time for an Object to Fall from a Given Height

[oldspice1212]

Hey guys so I have a quick question about the algebra for a question like this.
I personally like to do the algebra first before I plug in the numbers so I have question for this one where it seems would be safer to plug in the numbers first?

So won't be to tough,

A book falls from a shelf that is 1.75m above the floor. How long will it take the book to reach the floor?

So we have initial velocity = 0
Displacement = 1.75m
Acceleration/ gravity = 9.81m/s
Time=?

Its obvious that we can use the formula d=Vi(t)-1/2at^2

So my question here is for the algebra if we plug the numbers in first it would be easier since we have initial velocity =0 so we can do (0)t=0. But if we didn't this is where I have trouble since I'm not the best at algebra, we'd get

Square root( (2d/g)) - Vi(t) = t which confuses me because of the initial velocity * Time? What would we have to do with the time there.

Thanks, I hope I didn't confuse anyone lol.

 

[gneill]

You should end up with a quadratic equation which you solve accordingly.

Given your equation, d=Vi(t)-1/2at^2, you seem to be mixing up your coordinate directions. If the book is falling and you're taking the eventual displacement as positive (d = 1.75m), then your assumption is that "down" is positive. But then you write -1/2at² which appears to assume that "up" is positive (given a positive value for acceleration).

So. Let's say that "up" is positive, and the floor is the zero reference. Then the initial displacement is d and the final displacement is 0. Then you can write:

$0 = d + v_it - \frac{1}{2}at^2$ {where a is a positive constant, the magnitude of the acceleration}

There's your quadratic equation.

Alternatively, you could take the zero reference to be at the book's initial position on the shelf, and "down" to be positive. Then you might write:

$-d = 0 + v_it + \frac{1}{2}at^2$

In either case, if $v_i$ is zero then the quadratic becomes easy to solve...

 

[TSny]

If Vi ≠ 0, then you are dealing with a general quadratic equation in t. Use the quadratic formula to solve it. See here.

For a video review, try here.

[EDIT: I see gneill already provided an answer.]

 

[Chestermiller]

Hey guys so I have a quick question about the algebra for a question like this.
I personally like to do the algebra first before I plug in the numbers so I have question for this one where it seems would be safer to plug in the numbers first?

So won't be to tough,

A book falls from a shelf that is 1.75m above the floor. How long will it take the book to reach the floor?

So we have initial velocity = 0
Displacement = 1.75m
Acceleration/ gravity = 9.81m/s
Time=?

Its obvious that we can use the formula d=Vi(t)-1/2at^2

So my question here is for the algebra if we plug the numbers in first it would be easier since we have initial velocity =0 so we can do (0)t=0. But if we didn't this is where I have trouble since I'm not the best at algebra, we'd get

Square root( (2d/g)) - Vi(t) = t which confuses me because of the initial velocity * Time? What would we have to do with the time there.

Thanks, I hope I didn't confuse anyone lol.

So, if the initial velocity is zero, you can just substitute Vi = 0 into your final equation and get the correct result. But seriously, for arbitrary initial velocity, if you want to solve it algebraically, it is best to follow the advice of the previous posters.

 

[oldspice1212]

Ah quadratic formula, been ages since I've used that, thanks a lot guys, it helped :)!