Calculating Time in Air for a Launched Ball

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SUMMARY

The discussion centers on calculating the total time a ball remains in the air when launched vertically with an initial speed of 37.5 m/s. The correct approach involves recognizing that the time to reach the peak of the trajectory is 3.83 seconds, and the total time in the air is double that, resulting in 7.66 seconds. The kinematic equations, particularly vf = vi + at and d = vft - 1/2at², are essential for solving the problem accurately.

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Homework Statement


A ball is launched vertically up from the ground. The initial speed of the ball is 37.5m/s. How long does it stay in the air?

Homework Equations


The 5 kinematic equations

The Attempt at a Solution


I figured the final speed would be 0m/s, and used the formula vf=vi+a*t so 0=37.5+(-9.8)t
and got t=3.83s... I don't think that's right, and was wondering if someone could please help me out? Any help is appreciated. Thanks
 
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helpme2012 said:

Homework Statement


A ball is launched vertically up from the ground. The initial speed of the ball is 37.5m/s. How long does it stay in the air?


Homework Equations


The 5 kinematic equations


The Attempt at a Solution


I figured the final speed would be 0m/s, and used the formula vf=vi+a*t so 0=37.5+(-9.8)t
and got t=3.83s... I don't think that's right, and was wondering if someone could please help me out? Any help is appreciated. Thanks
Welcome to PF! The speed of the ball is 0 when it reaches the top of its trajectory, so the time you have calculated is the time it takes to reach the top of its path. What do you suppose is the time it takes to come back down to the ground?
 


PhanthomJay said:
Welcome to PF! The speed of the ball is 0 when it reaches the top of its trajectory, so the time you have calculated is the time it takes to reach the top of its path. What do you suppose is the time it takes to come back down to the ground?

Thanks for the reply, I appreciate being welcomed here. I'm really confused, I'm sorry but I honestly have no idea. I was thinking it would be another 3.83s to come back down but that can't be right..
 


helpme2012 said:
Thanks for the reply, I appreciate being welcomed here. I'm really confused, I'm sorry but I honestly have no idea. I was thinking it would be another 3.83s to come back down but that can't be right..
And why not?
 


PhanthomJay said:
And why not?

I'm just doubting myself right now haha, according to the book it's worth 4 marks so I figured maybe I was missing something.
 


helpme2012 said:
Thanks for the reply, I appreciate being welcomed here. I'm really confused, I'm sorry but I honestly have no idea. I was thinking it would be another 3.83s to come back down but that can't be right..

A ball is launched vertically up from the ground. The initial speed of the ball is 37.5m/s. How long does it stay in the air?

If you're stuck on a question, just always state the given even if you have no idea what to do!

Given: a = -9.81ms-2 (I prefer to write it like this on one line, one ms-2 is equal to a m/s/s or a m/s2

vi = 37.5ms-1

d = ?

vf = 0 (at the top of its trajectory, the ball stops momentarily, then comes back down)

vf2 = vi2 + 2ad

Find the displacement, and then you can use a different equation to solve for time. (You now have four variables. Before, the three variables you had in your given restricted you to the use of the only equation that does not have time. Solving for displacement opens up the possibility to more equations, four of which include time)
I had a feeling I just undermined what phantom said, so attempt the question again before you look at the spoiler please :)

Pick one of those equations and solve for t.

Hint: I would pick d = vft - 1/2at2 this way, the final velocity of zero cancels out the vt term entirely, and you can solve without using the quadratic equation.
 
Last edited:


Try using one of the other kinematic equations, the one with displacement as a function of initial velocity, time and acceleration. The initial velocity and the acceleration of gravity are known, and what it is the value of the displacement between its initial position when it is thrown from the ground and its final position when it comes back down to the ground? Solve for t...
 


Hey guys, I found the displacement to be 71.75m. I used the equation suggested by Fifty and did this:
d = vft - 1/2at2
71.75 = 0t - 1/2(-9.8)t2
71.75 = 4.9t2

I divided 4.92 from both sides so after I had this:

14.64 = t2

I took the square root of 14.64 and got 3.826s, rounded off to 3.83s

What do you guy think? Did I make any mistakes?
 


helpme2012 said:
Hey guys, I found the displacement to be 71.75m. I used the equation suggested by Fifty and did this:
d = vft - 1/2at2
71.75 = 0t - 1/2(-9.8)t2
71.75 = 4.9t2

I divided 4.92 from both sides so after I had this:

14.64 = t2

I took the square root of 14.64 and got 3.826s, rounded off to 3.83s

What do you guy think? Did I make any mistakes?
It depends upon what your final answer is for the total time of the flight from start to finish. Incidentally, watch your signs...if displacement is positive down, then acceleration of gravity is positive down...
 
  • #10


PhanthomJay said:
It depends upon what your final answer is for the total time of the flight from start to finish. Incidentally, watch your signs...if displacement is positive down, then acceleration of gravity is positive down...

7.66 seconds for the total flight from start to finish?
 
  • #11


helpme2012 said:
7.66 seconds for the total flight from start to finish?
Yes!
 

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