Calculating Time in Air for a Launched Ball

In summary: Very good!In summary, the ball launched vertically from the ground with an initial speed of 37.5m/s will stay in the air for a total of 7.66 seconds, with 3.83 seconds to reach the top of its trajectory and another 3.83 seconds to come back down to the ground. This was calculated using the 5 kinematic equations, specifically the equations for final speed and displacement.
  • #1
helpme2012
11
0

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
 
Physics news on Phys.org
  • #2


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?
 
  • #3


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..
 
  • #4


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?
 
  • #5


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.
 
  • #6


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:
  • #7


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...
 
  • #8


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?
 
  • #9


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!
 

What is the formula for calculating the time in air for a launched ball?

The formula for calculating the time in air for a launched ball is t = 2 * v * sin(theta) / g, where t is the time, v is the initial velocity, theta is the launch angle, and g is the acceleration due to gravity.

How do you determine the initial velocity of a launched ball?

The initial velocity of a launched ball can be determined by measuring the distance the ball travels and the time it takes to travel that distance. The formula for initial velocity is v = d / t, where v is the initial velocity, d is the distance, and t is the time.

What is the launch angle and how does it affect the time in air?

The launch angle is the angle at which the ball is launched from the ground. It affects the time in air because it determines the vertical and horizontal components of the initial velocity. A higher launch angle will result in a longer time in air, while a lower launch angle will result in a shorter time in air.

How does the acceleration due to gravity affect the time in air?

The acceleration due to gravity is a constant force that causes objects to accelerate towards the ground. It affects the time in air because it determines how quickly the ball will fall back to the ground. A higher acceleration due to gravity will result in a shorter time in air, while a lower acceleration due to gravity will result in a longer time in air.

What are some factors that can affect the accuracy of the calculated time in air?

The accuracy of the calculated time in air can be affected by factors such as air resistance, the shape and weight of the ball, and external forces like wind. The initial velocity and launch angle should also be measured as accurately as possible to ensure a more precise calculation.

Similar threads

  • Introductory Physics Homework Help
2
Replies
38
Views
1K
  • Introductory Physics Homework Help
Replies
14
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
905
  • Introductory Physics Homework Help
Replies
1
Views
684
  • Introductory Physics Homework Help
2
Replies
36
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
3K
  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
2K
Back
Top