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A ball is launched vertically up from the ground, how long does it stay in the air?

by helpme2012
Tags: ball, ground, launched, stay, vertically
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helpme2012
#1
Nov5-12, 09:55 PM
P: 11
1. The problem statement, all variables and given/known data
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?


2. Relevant equations
The 5 kinematic equations


3. 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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PhanthomJay
#2
Nov5-12, 10:01 PM
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Quote Quote by helpme2012 View Post
1. The problem statement, all variables and given/known data
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?


2. Relevant equations
The 5 kinematic equations


3. 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?
helpme2012
#3
Nov5-12, 10:06 PM
P: 11
Quote Quote by PhanthomJay View Post
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..

PhanthomJay
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Nov5-12, 10:15 PM
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A ball is launched vertically up from the ground, how long does it stay in the air?

Quote Quote by helpme2012 View Post
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?
helpme2012
#5
Nov5-12, 10:20 PM
P: 11
Quote Quote by PhanthomJay View Post
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.
Fifty
#6
Nov5-12, 10:23 PM
P: 35
Quote Quote by helpme2012 View Post
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 :)

Spoiler

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.
PhanthomJay
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Nov5-12, 10:27 PM
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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.....
helpme2012
#8
Nov5-12, 10:41 PM
P: 11
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?
PhanthomJay
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Nov5-12, 11:06 PM
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Quote Quote by helpme2012 View Post
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...
helpme2012
#10
Nov5-12, 11:25 PM
P: 11
Quote Quote by PhanthomJay View Post
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?
PhanthomJay
#11
Nov6-12, 10:12 AM
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Quote Quote by helpme2012 View Post
7.66 seconds for the total flight from start to finish?
Yes!


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