Simple kinematics problem I can't figure out!

1. Oct 7, 2015

MRi04

1. The problem statement, all variables and given/known data
A ball is thrown down with an initial velocity of 2.0 m/s from a height of 1.75 m, determine how long until it hits the ground.

initial velocity = 2 m/s
distance = 1.75 m
acceleration = 9.8 m/s2
time = ?

2. Relevant equations
d=vit+1/2a(t)2

3. The attempt at a solution
1.75=2t + 1/2(9.8)t2
1.75=2t + 4.9t2
Square root of both sides???
1.32 = 1.4t + 2.21t
1.32 = 3.61t
All divided by 3.61???
t = 0.36
.... I know the solution to be 0.43 seconds, but can't get there!

Any help is greatly appreciated!

2. Oct 7, 2015

RUber

That is where you went wrong...
Solve this as a quadratic equation.

3. Oct 7, 2015

Ellispson

In your solution,you took square root of both sides which is incorrect.Instead,solve the quadratic equation formed in t.You will get the time taken.

4. Oct 7, 2015

MRi04

Thanks RUber and Ellispson, really appreciate it, can you give any more detail, I'm not sure how to do that?

5. Oct 7, 2015

RUber

If you have an equation of the form : At^2 + Bt + C = 0, then
$t = \frac{ -B \pm \sqrt{B^2 - 4AC}}{2A}$
Only one of the solutions will be a positive time...choose that one.

6. Oct 7, 2015

MRi04

but what does C represent in the quadratic equation? Sorry, it has been a long time since I have used the quadratic equation.

7. Oct 7, 2015

MRi04

I mean, what does it represent in terms of the problem I posted?

8. Oct 7, 2015

RUber

You have 1.75 = 2t + 4.9t^2 . Can you put that into the form of At^2 + Bt + C = 0? C is the number that is not multiplied by t or t^2...I get -1.75.

9. Oct 7, 2015

Ellispson

The constant term.Which means the term not having the variable t in it.

10. Oct 7, 2015

MRi04

Oh of course, I see. I'll try that now

11. Oct 7, 2015

MRi04

That worked for me! Thanks very much to both of you. Happy physics-ing!