Find height, range and time of flight for projectile motion

  • Thread starter Robb
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  • #1
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Homework Statement


upload_2015-10-10_9-48-29.png


Homework Equations


kinematics.
a=F/m

The Attempt at a Solution


100cos(70)=342
100sin(70)=393.7
X=1/2(-9.8)t^2+100t; t=0. t=20.4
 

Answers and Replies

  • #2
SteamKing
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Homework Statement


See attached.

Homework Equations


Kinematic equations.

The Attempt at a Solution


100cos(70)=342 (x component)
100sin(70)=393.7 (y component)
You might want to check the arithmetic above before going further in this calculation. Also, don't forget to indicate units.

X=1/2(-9.8)t^2 + 100t = -4.9t^2 + 100t; t=0, t=20.4
V(final)= V(initial) + at =100 + (-9.8)(20.4)=-99.9m/s

I'm convinced t is not equal to 20.4.
 
  • #3
CWatters
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Ok so you have an equation for the initial vertical velocity. What else do you know about the motion in the vertical plane? Final velocity at the top = ? Acceleration =? What other equations of motion do you know that might be relevant?
 
  • #4
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y component = 94
velocity at the top = 0
acceleration from g = -9.8m/s

I know my basic kinematics and a=f/m
 
  • #5
gneill
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The problem states that the projectile is launched from the top of a building, yet doesn't state the height of that building. So what interpretation are we to place upon the Range and Time of Flight parts? Is a symbolic answer with an assumed starting height (say h) desired, or do we consider the landing elevation to be the same as the launch elevation? Is there any point to the building?
 
  • #6
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I assume the range to be where the projectile lands (on the ground) and the time would stop at that point as well. I guess there is no point to the building although the elevation effects time and distance traveled.
 
  • #7
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h=[v(initial)^2*sin(theta)^2]/2g = [(100)^2(sin70)^2]/-19.6 = -450.5

Not sure how this can be negative unless it is accounting for the height of the building?
 
  • #8
gneill
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h=[v(initial)^2*sin(theta)^2]/2g = [(100)^2(sin70)^2]/-19.6 = -450.5

Not sure how this can be negative unless it is accounting for the height of the building?
The g in this case is just the constant g without sign. You might try deriving the expression for the maximum height to see why this is.
 

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