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Simple projections

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  • #1
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A particle is projected with velocity vector (10i + 15j)ms^-1 where i and j are unit vectors in the horizontal and upward vertical directions respectively. Find its velocity vector 2 seconds later and its distance from the point of projection at this time.

Well first of all, could anyone explain the difference between the velocity vector after 2 seconds, and the position vector after 2 seconds?


working: I got 20i + 10.4j for the position vector after 2 seconds, using pythagoras I was able to find the distance from the point of projection (22.5m), for the velocity vector I didn't really know what to do, so I looked at the answer and deduced how they got it. So using the initial velocity vector (10i + 15j) as a point O, and the new position (20i + 10.4j) I manage to get the velocity vector to be (20-10)i + (15-10.4)j so 10i + 4.6j, however the answer is 10i - 4.6j

Am I wrong, or is the book? If further proof of working is needed I'll scan the paper I used.

Thanks.
 

Answers and Replies

  • #2
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You can't mix velocity and position vectors in the same equation freely. The units wouldn't make sense.

This is a particle under the influence of gravity, is it not? Every second, the particle is losing vertical velocity but not horizontal velocity. How much vertical velocity must it lose every second? How much vertical velocity does it lose after two seconds?
 
  • #3
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Sorry to say but u are wrong

ok do u know all about the projectile motion??

in projectile horizontal component(here 10i) dosent change.
So initial and final velocity of projectile along that direction will be same

now for vertical direction, there is acceleration g=9.8 m/s^2

so using Final velocity= initial velocity + Acceleration*Time we get...

v(along j axis)= 15(initial)-9.8*2 {acceleration is downward but projection is upward so a =-9.8}

Solving we get v(along j axis) = -4.6

so answer is 10i^ -4.6j^
 
  • #4
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my bad, vertically it loses v = 15 - 9.8*2 = -4.6, and horizontally it gains 10*2 = 20

thanks!
 
  • #5
CAF123
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The position vector can be described as a vector from the origin of an assigned coordinate system to the point of the particle in the xy plane, defined by it's x and y coordinates (in this case). This position vector will continually change as the particle moves along some trajectory. The velocity vector is always tangential to the particle's trajectory.
So we know [itex] v_o = 10\hat{i} + 15\hat{j}. [/itex] I assume we are in an environment with no air resistance, so the horizontal component is fixed for the duration of the flight (as can be verified by calculation).
The vertical component of velocity at some time [itex] t [/itex] is given by [itex] v_y = v_{oy} -gt. [/itex] Use this to compute your vertical component after 2 seconds.
 
  • #6
54
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........horizontally it gains 10*2 = 20

No there is no acceleration along horizontal so horizontal component remains same
 

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