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Vector problems

  1. Sep 20, 2008 #1
    1. The problem statement, all variables and given/known data
    A cannon located 60m from the base of a vertical 25 m cliff shoots a shell at 43 degrees above the horizontal.

    What is the minimum muzzle velocity be for the shell to clear the top of the cliff ?
    The ground at the top of the cliff is level, with constant elevation of 25 m above the cannon, How far does the shell land past the cliff ?
    __________________________________________________ ________________
    A man stands on the roof of a 15 m tall building a throws a rock with a velocity of magnitude 30 m/s at an angle of 33 degrees above the horizontal.

    What is the maximum height above the roof reached by the rock .?
    The magnitude of the velocity of the rock before it strikes the ground ?
    The horizontal distance from the base of the building to the point when the rock strikes the ground .?
    __________________________________________________ ________________________
    A ball is thrown upward with an initial spped of 20 m/s from the edge of a 45 m high cliff. At the istant the ball is thrown, a woman starts running away from the base of the cliff with a constant speed of 6 m/s .
    At what angle above the horizontal should the ball be thrown so the runner will catch it just before it hits the ground and how far does the women wrun before she catches the ball .?


    2. Relevant equations
    Vy=vsin
    Vx=vcos
    dx= vx * t
    dy = Vyt - .5at^2

    3. The attempt at a solution
    First problem I do not know where to start.

    Second problem I tried and got the answers wrong to all 3. I first solved the velocities for the x and y directions. Then using Vy found out how long it took to hit the ground then use half of that to find the maximum height. But i got that wrong and the rest followed.

    Third problem is used 6 m/s = Vx and then figured out the angle using Vx=Vcos then figured out Vy then I was stuck.
     
  2. jcsd
  3. Sep 20, 2008 #2
    For the second problem, think about what the problem is saying. The man is throwing a rock up a certain height. You know that at the maximum height, the final velocity will be 0. You know that acceleration due to gravity is -9.81m/s^2, and you know the initial Vy. You should be able to solve for how far up the rock is traveling. This is a unique projectile motion problem in that the time it takes to reach maximum height is NOT the same as the time it takes to fall from that height and hit the ground.

    For the third problem, think about what needs to be true about the Vx of the ball if the woman running below at 6 m/s is going to catch it. From there, hopefully you can figure out the rest.

    If this doesn't help you, I'll try to explain it differently.
     
  4. Sep 20, 2008 #3
    for the second problem i found maximum height:
    vf^2=vi^2+2ad
    which: vi=vy=16.3m/s a= -9.81 m/s^2
    d=13.6 m
    Im stuck on these now
    The magnitude of the velocity of the rock before it strikes the ground ?
    The horizontal distance from the base of the building to the point when the rock strikes the ground .?
     
  5. Sep 20, 2008 #4
    Assuming air drag is nonexistent, the horizontal velocity will not change. The vertical velocity can now be thought of a "dropping an object off a cliff" problem where the height of the "cliff" would be the height of the building plus the height you calculated in part a.

    For part c, there is no acceleration so you will only need to use the equation S = VT where S is the horizontal distance from the base of the building. I hope this helps you out a little bit!
     
  6. Sep 20, 2008 #5
    okaii i got c
    for b thought you will use the equation
    vf^2 = vi^2 +2ad
    vi = 0
    d = (answer to a) + 15m
    a = +9.81 m/s^2

    solve for vf and thats the answer
    is that correct.?
     
  7. Sep 20, 2008 #6
    Yes that's correct :) I'm glad I was able to be of some assistance. Please let me know if there is anything else I can do to help you.
     
  8. Sep 20, 2008 #7
    A cannon located 60m from the base of a vertical 25 m cliff shoots a shell at 43 degrees above the horizontal.

    What is the minimum muzzle velocity be for the shell to clear the top of the cliff ?
    The ground at the top of the cliff is level, with constant elevation of 25 m above the cannon, How far does the shell land past the cliff ?
     
  9. Sep 20, 2008 #8
    Separate the Initial Velocity in x and y components.

    Find Vyi

    You know the Vf is 0, the height, and the acceleration due to gravity.

    Once you have the Initial Vy, you can use trig to solve for the minimum muzzle velocity.

    After you have solved for this, you can solve for the Vx component and then use S = VT to solve for part b.

    Hope I gave you enough hints ;)
     
  10. Sep 20, 2008 #9
    but how can i find Vy if i dont know V and just the degrees
     
  11. Sep 20, 2008 #10
    Part a wants you to solve for V, which you can do as long as you have the angle (which you do) and Vy.

    As I said before, you said looking for a vertical velocity which can reach a maximum height of 25 meters if acceleration due to gravity is -9.81m/s^2.

    Does this help?
     
  12. Sep 20, 2008 #11
    so you can do Vf^2= Vi^2 +2ad

    which vf=0
    d=25
    a= -9.81

    this gives you vy and then you can find v ?
     
  13. Sep 20, 2008 #12
    Correct. You're on the right track.
     
  14. Sep 20, 2008 #13
    that would be the answer for part a .?

    then for b i use v to find vx then but i dont know either d or t for d=vx * t
     
  15. Sep 20, 2008 #14
    So then try to calculate t. Think about how long this object will be in the air. Try drawing a picture if that helps.
     
  16. Sep 20, 2008 #15
    do d = vy*t +.5at^2

    and d = 25 and i know vy and a

    ?
     
  17. Sep 20, 2008 #16
    A much easier way would be to use v = vi + at to solve for time.

    But that time will not be the total time. Why not? Again, think about what that time represents.
     
  18. Sep 20, 2008 #17
    how long it takese to get half way up..... so would you double time once yuh find it ?
     
  19. Sep 20, 2008 #18
    No because it's not falling back down another 25 meters. Remember, you're firing an object towards the top of a cliff. At the maximum height, the object will just be over the cliff and then it will fall down and land on the ground at the top of the cliff.

    Try to figure out how high above the top of the cliff the object will be at v = 0. Once you have this, you can figure out how much longer the object stays in the air.
     
  20. Sep 20, 2008 #19
    so Vf^2=Vy^2+2ad
    which vf = 0
     
  21. Sep 21, 2008 #20
    I apologize. When I went back and re-read the problem, I discovered that I had missed a step.

    At the beginning, when you calculate Vy, you understand that the corresponding Vx must be large enough to reach the cliff, which is 60 meters away.

    I calculated the time the projectile took to travel the 60 meters (Sx = VxT) and realized that that must be equal to the time at which the projectile reaches maximum height.

    Try doing this and then resolving for velocity.

    Sorry again.
     
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