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Acceleration due to gravity

  • Thread starter eoneil
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  • #1
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Homework Statement


A particle moves from rest with a horizontal velocity of 30 meters per second, and is subject to a vertical uniform acceleration of 9.81 meters/sec2 downwards. Find its distance from the starting point at the end of ten seconds.


Homework Equations


vf= at+vi
t=s/v where t= time, s= distance, v= velocity

The Attempt at a Solution


The mass of the object is not given, which one would assume means it is irrelevant, as all objects fall under the constant of 9.81m/s2. Since the particle moves from rest to 30m/s, Vinitial=0. The problem must be separated into its constituents, what is occurring on the y axis, and x axis. However, nothing is given about whether the particle begins moving at an angle, or whether it initiates movement from an elevated position.
I'm stuck here. Any suggestions?
 

Answers and Replies

  • #2
1,137
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Are you familiar with the fact that movement in X axis is independent of movement in Y axis?

and vice versa
 
  • #3
danago
Gold Member
1,122
4

Homework Statement


A particle moves from rest with a horizontal velocity of 30 meters per second, and is subject to a vertical uniform acceleration of 9.81 meters/sec2 downwards. Find its distance from the starting point at the end of ten seconds.


Homework Equations


vf= at+vi
t=s/v where t= time, s= distance, v= velocity

The Attempt at a Solution


The mass of the object is not given, which one would assume means it is irrelevant, as all objects fall under the constant of 9.81m/s2. Since the particle moves from rest to 30m/s, Vinitial=0. The problem must be separated into its constituents, what is occurring on the y axis, and x axis. However, nothing is given about whether the particle begins moving at an angle, or whether it initiates movement from an elevated position.
I'm stuck here. Any suggestions?
I think you can assume that the initial vertical velocity is zero, so it only moves off from rest with a horizontal speed of 30 m/s i.e. at an angle of 90 degrees to the vertical. As for its initial position, this is not relevant since you only need to find its distance with respect to the starting point
 
  • #4
1,137
0
No. i meant that forces acting along Y axis donot effect any thing that is along X or Z axis, be it speed , momentum acc. etc

you can take it as a wholly separate motion.

EDIT:

The body will move down but it has nothing to do with the X axis's speed, acc. ar anything. same applies for movement in X axis
 
  • #5
18
0
As its only horizontal velocity we just need to take into account the horizontal initial component, and the vertical and coefficient of restitution are irrelevant in this case. Is that what you're saying danago?

Solving for horizontal displacement at time:
∆x= vx0t, where vx0 is initial horizontal velocity, t is time.
∆x= (30m/s)(10s)= 300m at 10s.
But there is no initial velocity since it starts from rest. And why is gravity included if it has no relevance to the problem?
 
  • #6
1,137
0
Try making proper figure and understanding how the body is moving

You have just founded displacement along X ...

Now along Y
v = 0
a = g
So find distance travelled in Y direction ... Then Find distance b/w initial and final position
 
  • #7
18
0
Solving for vertical displacement, v = 0, a= g,
∆y= yx0t- ½ gt2
∆y= (0)(10s)- ½ (9.81m/s2)(10s)2= -490.5m

Now that displacement is solved on both axes, how do I obtain initial position? I know the particle, at 10s, is 300m ahead on the x axis, and -490.5m on y. Do I combine the two?

delta d = v delta t?
 
  • #8
1,137
0
Now you knw X and Y distance ...

particle's displacement is something like on the hypotenuse of right angle triangle. and you know the two sides of triangle ...
how do you find hypotenuse??
 
  • #9
18
0
pythagoras' theorem,

so 300^2+(-490.5^2)= c^2
c= 575m
 
  • #10
1,137
0
Yes ... is that right answer?
 

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