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Constant Acceleration Question

  • Thread starter habs.fan
  • Start date
11
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If an 81.00kg projectile is fired STRAIGHT up, with no air resistance, under the effects of gravity, and it reaches an altitude of 175km (175,000 metres), what was it's initial speed (what speed was it fired upwards at)?
I disregarded the objects mass and calculated 1,715,000 m/s...would these be correct?
Also, how would I create a formula for speed vs. time? I know it would be a quadratic...probably -4.9t^2, or something along those lines?
 

Answers and Replies

Doc Al
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I disregarded the objects mass and calculated 1,715,000 m/s...would these be correct?
How did you arrive at this number?

(Moved to Intro Physics forum.)
 
11
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To be honest I've forgotten now, either d = vit + (0.5)at2 or a = (vf-vi) / t
 
27
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I calculated 1,715,000 m/s...would these be correct?
Ummm... Just a little out, that's over half the speed of light.

You were correct to disregard the mass. Firstly i would suggest writing down all the information you know.

Acceleration :
Displacement:
Initial Velocity:
Time:
End Velocity:
(One of these isn't relevant in this example)

After you have your values substitute it into the equation of motion with all the relevant pieces of information.
 
Last edited:
11
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Acceleration = -9.8m/s/s
Displacement = 175 000 metres
Initial Velocity = Unknown
Time = Unknown
End Velocity = 0 (assuming it comes back down and lands?)

So to find Vi, I would do: Vf^2 = Vi + 2ad
0^2 = Vi^2 + 2(-9.8)(175,000)
Correct?
 
Last edited:
27
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Yes all is correct expect your equation it's :

v^2 = u^2 + 2as or as you would write Vf^2 = Vi^2 + 2ad

End Velocity = 0 (assuming it comes back down and lands?)
Correct?
I think you are confused, we are talking about only the first half of the journey, from the ground to the point where the object is at the point of coming back down (Has zero velocity.)
 
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Alright so for the first part of the question, I get a new and improved answer of 1852m/s, which is much more reasonable.
Now for the second part, I'm thinking I should make a quadratic function out of this...would I use 1/2(a)(t)^2 for this? Or...is there an easier way to solve it?
 
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Yeah that looks good. It asks you for a quadratic equation and the only equation of motion that contains a square of t is :
s = ut + 0.5at^2
 
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Actually wait a minute ignore what i said. I assumed because you assumed the equation would be quadratic that it would be... if that makes any sense?

You want an equation of speed vs time so you want time and speed to remain as symbols t and v.
 
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Except I'm actually looking for when it's displacement is > 100,000 metres, so I need a displacement vs. time graph?
 
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[tex]v_y=v_{0y}+gt[/tex]

[tex]y-y_0=\frac{v_{0y}+v_y}{2}\times{t}[/tex]

solve for t, cross out things you're not given and solve for [tex]v_{0y}[/tex]
 
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247 Seconds. DONE. Thanks for all of your help, fellow physics fans :)
 

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