Projectile Motion Analysis: Time in Air Calculation

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SUMMARY

The discussion focuses on calculating the time a projectile remains in the air when fired horizontally from a height of 45.0 meters with an initial speed of 250 m/s. The correct approach involves recognizing that the horizontal velocity is extraneous to the vertical motion, which is governed solely by gravity. The time in the air is calculated using the equation for free fall, yielding a result of 3.03 seconds. Additionally, the horizontal distance traveled during this time is calculated to be approximately 758 meters.

PREREQUISITES
  • Understanding of kinematic equations, specifically for free fall.
  • Familiarity with projectile motion concepts.
  • Knowledge of basic calculus principles as they apply to physics.
  • Ability to differentiate between horizontal and vertical components of motion.
NEXT STEPS
  • Study the derivation and application of the kinematic equation y = Vo*t + 0.5*a*t^2.
  • Learn about the effects of air resistance on projectile motion.
  • Explore the relationship between launch angle and range in projectile motion.
  • Investigate the concept of maximum height and its calculation in projectile motion.
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the principles of projectile motion and free fall calculations.

Blox_Nitrates
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Homework Statement


A projectile is fired horizontally from a gun that is 45.0m above flat ground, emerging from the gun with a speed of 250m/s. How long does the projectile remain in the air?


Homework Equations


unsure of what equation to use.


The Attempt at a Solution


I tried to use t = Δx/Vox which didn't work because it gave me 45/250=.18

*The answer is 3.03 seconds in the back of the book.
 
Last edited:
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Blox_Nitrates said:

Homework Statement


A projectile is fired horizontally from a gun that is 45.0m above flat ground, emerging from the gun with a speed of 250m/s. How long does the projectile remain in the air?


Homework Equations


unsure of what equation to use.


The Attempt at a Solution


I tried to use t = Δx/Vox which didn't work because it gave me 45/250=.18
If you want to use that eqn to find t then you will need the horizontal distance that the body travels. What other kinematic equations do you know?
 
I have these
v=vo+at
x-xo=volt+(1/2)at^2
v^2=Vo^2+2a(x-xo)
x-xo=(1/2)(vo+v)t
x-xo=vt-(1/2)at^2

Sorry about asking something probably simple as to I'm new to Calculus based Phyics and have never seen physics before.
 
Question: If you dropped a bomb from a stationary hot air balloon while simultaneously dropping another bomb from a horizontally flying aircraft at the same altitude, would there be any difference in the amount of time it takes for the bombs to hit the ground?
 
LawrenceC said:
Question: If you dropped a bomb from a stationary hot air balloon while simultaneously dropping another bomb from a horizontally flying aircraft at the same altitude, would there be any difference in the amount of time it takes for the bombs to hit the ground?

No, there wouldn't be.
 
That is correct. Therefore you have a free fall problem. The muzzle velocity is analogous to the moving airplane. It is extraneous data that is supplied to confuse the student.
 
LawrenceC said:
That is correct. Therefore you have a free fall problem. The muzzle velocity is analogous to the moving airplane. It is extraneous data that is supplied to confuse the student.

So in this case t = (V-Vo)/a = ((250m/s)/-9.8m/s^2)?

but a = 0 due to no acceleration because of horizontal motion.
 
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No, the 250 m/s is the horizontal velocity and is extraneous data. The problem is the same as if you dropped a brick off a 45 m high building. How long would it take to reach the ground?
 
t = (0-45m/s)/-9.8m/s^2 = 4.59s?

Is that correct?
 
  • #10
You are attempting to subtract meters/second from meters. You cannot do that.

What about the formula

y = Vo*t + .5*a*t^2

where Vo is initial VERTICAL velocity
a is acceleration of gravity
t is time
y is distance of free fall
 
  • #11
45=0*t+.5(9.8)t^2

(45/4.9)^(1/2) = 3.03s

Thank you so much for your help LawrenceC!

So if I was asked At what horizontal distance from the firing point does it strike the ground? I would use a similar eqn to solve it while plugging in t correct?
 
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  • #12
The horizontal velocity does not change (neglecting air drag), so you can use your initial formula (in post 1) to calculate that distance.
 
  • #13
mfb said:
The horizontal velocity does not change (neglecting air drag), so you can use your initial formula (in post 1) to calculate that distance.

I'm not sure if this is correct but I used Savg=TotalDistance/Δt
Total Distance = 250m/s(3.03s) = 758m rounded up.
The answer does match the one in the book.
 
  • #14
That is correct.
 
  • #15
This question throws me off.
A projectile's launch speed is five times its speed at maximum height. Find launch angle θo.
 
  • #16
That is a nice question.

At maximum height, can you determine the direction of motion? Can you say something about the launch velocity components, based on that?
 
  • #17
I can say that the max height will be vertical and the direction of motion is upwards. Wouldn't that be a 90 degree angle of (pi/2)?
 
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  • #18
and the direction of motion is upwards
If it would go upwards there, it would not have reached its maximal height.

Wouldn't that be a 90 degree angle of (pie/2)?
It is pi, pie is something to eat. And it depends on the definition of the angle.
 
  • #19
mfb said:
If it would go upwards there, it would not have reached its maximal height.


It is pi, pie is something to eat. And it depends on the definition of the angle.

Sorry. So would the maximal height be somwhere between 45-90 degrees?
 
  • #20
Blox_Nitrates said:
Sorry. So would the maximal height be somwhere between 45-90 degrees?

It doesn't mean that the angle is chosen to achieve maximum height. The angle is whatever it is. mfb is asking what the direction of motion of the projectile will be when the projectile is at its maximum height on the trajectory. Sketching it might help.
 
  • #21
haruspex said:
It doesn't mean that the angle is chosen to achieve maximum height. The angle is whatever it is. mfb is asking what the direction of motion of the projectile will be when the projectile is at its maximum height on the trajectory. Sketching it might help.

I would see a parabolic function in that case but I still feel like I'm not seeing something that I'm possibly missing.
 
  • #22
Blox_Nitrates said:
I would see a parabolic function in that case but I still feel like I'm not seeing something that I'm possibly missing.
That's right. So which way is the projectile moving when at max height?
 
  • #23
haruspex said:
That's right. So which way is the projectile moving when at max height?

Somewhat angled about 60 degrees moving upward, hitting the apex when a=0 and then moving downward towards the ground.
 
  • #24
Blox_Nitrates said:
Somewhat angled about 60 degrees moving upward,
If it's moving upward it's not yet at max height.
hitting the apex when a=0
That's max height of the trajectory. Which way is it moving there?
 
  • #25
haruspex said:
If it's moving upward it's not yet at max height.

That's max height of the trajectory. Which way is it moving there?

It would be moving horizontally for an instant before moving downwards.
 
  • #26
Blox_Nitrates said:
It would be moving horizontally
Yes. Now, suppose it has speed v there. What then can you say then about its components of velocity at launch?
 
  • #27
V=Vo+at?
 
  • #28
Blox_Nitrates said:
V=Vo+at?
What horizontal acceleration is there (if air resistance is ignored)?
 
  • #29
haruspex said:
What horizontal acceleration is there (if air resistance is ignored)?

There wouldn't be any acceleration right? But then that would leave V=Vo which is true?
 
  • #30
Blox_Nitrates said:
There wouldn't be any acceleration right? But then that would leave V=Vo which is true?
That depends what you mean by Vo. Is that the magnitude of the initial velocity, or of just a component of it?
 

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