# Time a ball is in the air.

1. Jan 25, 2015

### redroomreflex

1. A ball is thrown vertically upward with an initial velocity of 27.0 m/s. Neglecting air resistance, how long is the ball in the air?

2. Relevant equations

vf (v final) = v0 (v initial) + at

3. The attempt at a solution

I set final velocity equal 0. So..

0 = 27.0 m/s (initial velocity) + (-9.8 m/s2)t

-27.0 m/s divided by -9.8

and it turned out to be equal to 2.76 seconds.

What am I doing wrong here? I saw a similar thread and the problem was solved the same way; but apparently my answer is incorrect.

2. Jan 25, 2015

### Simon Bridge

Welcome to PF,
What makes you think the final velocity is zero?

Describe what happens to the speed of the ball as it travels - using words.
The ball is thrown at high speed, directly upwards ... is gets slower and slower until...

3. Jan 25, 2015

### redroomreflex

Thanks Simon.

And as dumb as this reasoning is... final velocity equals zero because that's when the ball is momentarily in the air for as it changes a downward direction down to the ground....

4. Jan 25, 2015

### Simon Bridge

The ball has to be instantaniously stationary when it turns around - yes.
Does the ball turn around at the end of it's flight through the air?

5. Jan 25, 2015

### redroomreflex

No, it doesn't. It goes through an upside down parabolic motion?.. yes?

6. Jan 25, 2015

### Simon Bridge

Well done - at what part of the motion is the speed zero then?
Is it a quarter the way through, three quarters? What?

This is where I encourage students to get a ball and toss it in the air ... watch carefully.
Everybody has seen this happen, but you usually have no reason to think about what you are seeing or really pay attention.
A large chunk of science is about really paying attention to what is right under your nose - it's rewarding too because it means you are fully experiencing the phenomena around you. This is called "living life to the fullest".

7. Jan 25, 2015

### redroomreflex

The part of the motion when the speed is zero is at the halfway point.

8. Jan 25, 2015

### Simon Bridge

Well done - so the time you calculated was for the ball to reach which point?
What fraction of the time the ball was in the air is this?

9. Jan 25, 2015

### redroomreflex

The time calculated was when the ball's velocity is momentarily zero at the halfway point... so the fraction of time the ball is in the air would be half that time?

10. Jan 25, 2015

### Simon Bridge

So the time you calculated was half the time the ball was in the air - so what was the time the ball was in the air?

11. Jan 25, 2015

### redroomreflex

1.38 seconds.

12. Jan 25, 2015

### Simon Bridge

So you are saying that if half the time the ball is in the air is 2.76 seconds, then the whole time the ball is in the air must be 1.38 seconds?

So you catch the ball 1.38s after you threw it, but it turned around 2.76s after you threw it?
Does that make sense?

13. Jan 25, 2015

### redroomreflex

Okay... no... so it would be 2.76 seconds that it is in the air.

14. Jan 25, 2015

### Simon Bridge

Your first calculation said that it took 2.76 seconds for the ball to reach zero velocity.
You have said, above, that it reaches zero velocity when it turns around.

So the order of events must be:
t=t0 you throw the ball
t=t0+2.76s the ball turns around

t=t0+? you catch the ball.

The question wants the time indicated by the question mark.
You have said that the ball turns around half way through the motion.
So if that is half way, what time is all the way?

Consider: if 300ml is half the glass, how many mls in the whole glass?

15. Jan 25, 2015

### redroomreflex

5.52 seconds.

16. Jan 25, 2015

### Simon Bridge

Well done - in maths, if the time you want is T, and you know that half the time is t, then t=T/2 so T=2t.

Where you went wrong in post #1 is you picked the wrong endpoint.
Since the velocity of the ball when you catch it is not zero - what is it? What velocity should you have chosen?