# Quadratic equation ball 72km/h

1. Jun 30, 2013

### LDC1972

1. The problem statement, all variables and given/known data

A ball is thrown at 72km/h speed from top of a building. Building is 125m tall.

2. Relevant equations

Distance travelled before it hit the ground is as follows:

s = Ut + 0.5 gt^2

3. The attempt at a solution

Using quadratic equation:
x = -b±b^2 - 4ac / 2a

The question is for part one:
Find the time for the ball to drop to fifth of the height of the building.

Now I'm not sure if that means to 25m or 100m?!
For 25m = 1 second
For 100m = 3.3852

I got these from the quadratic equation. Do they seem right?

Thanks!

2. Jun 30, 2013

### darkxponent

The question is not clear. Whether the ball is thrown up or thrown down?

This is wrong. There is square root missing.

3. Jun 30, 2013

### LDC1972

Sorry again I clicked the root sign but it couldn't of worked. Here is the equation again:

x = -b ± √b^2 - 4ac / 2a

The question is unclear on the "fifth" part. Do they mean from top or bottom?

Anyway I calculated it from all heights:

At 25m = 1 second
At 100m = 2.899 seconds
At 125m = 3.3852 seconds

Seems right?

4. Jun 30, 2013

### darkxponent

The question is clear here. They asking from the top that is 25m.

Should i solve it and check it? I am not going to do that. It would better if you tell me what you put the values of U, g and S in this equation. I think this would be enough to enough to tell whether you did it correctly or wrong.

s = Ut + 0.5 gt^2

5. Jun 30, 2013

### LDC1972

OK,
U = 20ms (metres per second) (initial velocity of 72 Km/h I converted to ms)
g = 10ms (this is given in the question (g = acceleration due to gravity 10ms). Without which I'd of put 9.81ms, but I assume they allowed for ball mass?)
s = distance, so 25 metres, 100 metres and 125 metres (although by the sounds of it I can drop the 100 metre one)

6. Jun 30, 2013

### LDC1972

Ah, of course! I got so wrapped up in the quadratics and factoring today I forgot I had the original equation to "plug" my t into and check the solution I got was correct :-)

7. Jun 30, 2013

### darkxponent

Everything looks so damn correct. Good work!

You can always press the thanks button if you find my help useful :)

8. Jun 30, 2013

### LDC1972

Done the thanks button. Brilliant mate THANK YOU!!!!!!!!!!!!!!!!!!!!!!

9. Jun 30, 2013

### darkxponent

As the question doesn't state whether the ball is thrown upwards or downwards. You have taken the downward case.

Question: How will the values change if the ball is thrown upwards?

10. Jun 30, 2013

### LDC1972

The question does say downwards, sorry for my poor copying!
I'm done for the day, but if it had been upwards g would of worked "against", not "for" the acceleration.

11. Jul 1, 2013

### LDC1972

12. Jul 1, 2013

### darkxponent

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