# 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!