Why is my time calculation for the ball's flight incorrect?

In summary, the ball is thrown vertically upward with an initial velocity of 27.0 m/s. Using the equation vf = vo + at, where vf is final velocity, vo is initial velocity, a is acceleration, and t is time, it can be determined that the ball will reach a height of 0 meters at the halfway point of its motion. Therefore, the time it takes for the ball to reach its maximum height and then return to the starting point is 2.76 seconds. However, the question is asking for the total time the ball is in the air, which is double the time it takes to reach the highest point. Therefore, the ball is in the air for a total of 5.52 seconds.
  • #1
redroomreflex
8
0
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?

Homework Equations



vf (v final) = v0 (v initial) + at[/B]

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.
 
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  • #2
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
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
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
No, it doesn't. It goes through an upside down parabolic motion?.. yes?
 
  • #6
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
The part of the motion when the speed is zero is at the halfway point.
 
  • #8
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
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
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
1.38 seconds.
 
  • #12
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
Okay... no... so it would be 2.76 seconds that it is in the air.
 
  • #14
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
5.52 seconds.
 
  • #16
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?
 

Related to Why is my time calculation for the ball's flight incorrect?

1. How do you calculate the time a ball spends in the air?

The time a ball spends in the air can be calculated using the formula t = √(2h/g), where t is the time, h is the initial height of the ball, and g is the acceleration due to gravity (9.8 m/s²).

2. Does the mass of the ball affect the time it spends in the air?

No, the mass of the ball does not affect the time it spends in the air. This is because the formula for calculating time in the air only takes into account the initial height and acceleration due to gravity.

3. How does air resistance affect the time a ball is in the air?

Air resistance has a minimal effect on the time a ball spends in the air. This is because the formula for calculating time in the air assumes no air resistance. In reality, air resistance may slightly decrease the time the ball is in the air.

4. Can the angle at which a ball is thrown affect the time it spends in the air?

Yes, the angle at which a ball is thrown can affect the time it spends in the air. The formula for calculating time in the air assumes the ball is thrown at a 90-degree angle, which results in the maximum time in the air. Throwing the ball at a different angle will result in a shorter time in the air.

5. How does the height from which a ball is thrown affect the time it spends in the air?

The initial height from which a ball is thrown directly affects the time it spends in the air. The higher the initial height, the longer the time in the air. This is because the formula for calculating time in the air includes the initial height as a variable.

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