Basic Physics Question please help

  • Thread starter kara
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  • #26
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you said i left out the vertical displacement at time, so that would be y in the case of this equation y = y0 + v0y*t - 1/2*g*t^2 , and y is 53 m ?
 
  • #27
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alright so solving that equation would give me v0. Now what do i do with that? would i also need to solve for v0 in equation x=x0 + v0t?
 
  • #28
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:O I just figured something out... Don't know if im on the right track but let me know:

If i solve for v0 in equation x=x0 + v0t, then i can sub that into the other equation?
 
  • #29
Doc Al
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Solving that equation gives you the vertical component of the initial velocity; you already solved for the horizontal component.
 
  • #30
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Alright, so i got 36.3 m/s for v0
 
  • #31
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So the horizontal component of my initial velocity is 20 m/s, and the vertical component of initial velocity is 36.6 m/s
 
  • #32
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Now i have to calculate how far is its displacement horizontally from launch pt, at the instant it achieves max height.

So i know that its v will be 0 m/s at max heigh.
 
  • #33
Doc Al
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kara said:
So i know that its v will be 0 m/s at max heigh.
Right... the vertical component of the velocity will be zero.
 
  • #34
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and b/c im looking for how far its been displaced horizontally from the launch pt. i am looking for x?
 
  • #35
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so i can solve for t in the y=y0 + v0-1/2gt^2 equation and sub t into x=x0 +v0t equation and solve for x
 
  • #36
Doc Al
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kara said:
and b/c im looking for how far its been displaced horizontally from the launch pt. i am looking for x?
That's right. You are looking for the value of x when y is maximum. Hint: When does it reach the maximum height?
 
  • #37
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well the max height is 53 m, and it reaches that height when v = 0.0 m/s
 
  • #38
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i plugged in all my values but get stuck at one point with a negative square root:

53 = -1/2(9.8)t^2
 
  • #39
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if i multiply both sides by 2 to get rid of the 1/2 i get

106 = -(9.8) t^2
 
  • #40
Doc Al
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kara said:
i plugged in all my values but get stuck at one point with a negative square root:

53 = -1/2(9.8)t^2
You left out part of that equation; it should be:
53 = v0t -1/2(9.8)t^2

Where v0 is the vertical component of initial velocity that you found earlier.
 

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