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Projectile Motion Problems Involving angles

  1. Sep 21, 2011 #1
    There are two problems involving projectile motion and angles I am having problems with
    1. The problem statement, all variables and given/known data

    You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics, you ask one of the archers to shoot an arrow parallel to the ground. You find the arrow stuck in the ground 61.0 away, making a 3.00 angle with the ground.

    x = 61m
    Theta = 3 degrees

    2. Relevant equations
    y = (1/2)a(t^2)
    y = xtan(theta)
    Vox = x/t

    3. The attempt at a solution
    I first used y = xtan(theta) and got y=(61)tan(3)=3.196m From there i used y=(1/2)a(t^2) to solve for t and got t=0.808s. from there i used Vox=(61)/(0.808) and got 74.5m/s. The webste claims this is wrong. Any help?


    Problem 2:
    1. The problem statement, all variables and given/known data
    A baseball player friend of yours wants to determine his pitching speed. You have him stand on a ledge and throw the ball horizontally from an elevation 6.00 above the ground. The ball lands 30.0 away. What is his pitching speed?

    I solved that part and got V=21.7m/s then there is the second part of the question.

    As you think about it, you’re not sure he threw the ball exactly horizontally. As you watch him throw, the pitches seem to vary from 5° below horizontal to 5° above horizontal. What is the range of speeds with which the ball might have left his hand? Enter the minimum and the maximum speed of the ball.

    y=6m
    x=30m
    V=21.7m/s?
    Theta=5 degree?

    2. Relevant equations
    y = Vyt + (1/2)a(t^2)
    Vy=Vtan(theta)
    Vox = x/t

    3. The attempt at a solution
    I honestly don't really know what to do for the second part. I tried to find the velocity for if he had thrown 5 degree above the parallel.
    So i got Vy=Vtan(5)=2.4
    Then i used Y=Vyt+(1/2)a(t^2)=2.4t+(-4.9)(t^2) and solved for t and got t=1.37s and then used V=x/t=30/1.37=21.9m/s

    I dont know if thats right because i have yet to press submit because i can not for the life of me figure out what the velocity would be if he threw the ball 5 degrees below the parallel
     
  2. jcsd
  3. Sep 21, 2011 #2

    rl.bhat

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    Homework Helper

    I first used y = xtan(theta)
    Th above statement is wrong. θ is the angle made by the tangent drawn at the point of impact arrow to the ground.
    tanθ = vertical velocity/ horizontal velocity.
    If t is the time of flight, x is the range of the projectile, find the vertical velocity and the horizontal velocities.
     
  4. Sep 21, 2011 #3
    How would I go about doing that?
     
  5. Sep 22, 2011 #4
    Okay i got the arrow one. can anyone help with the baseball one?
     
  6. Sep 22, 2011 #5

    rl.bhat

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    Homework Helper

    Refer second thread below. ( Projectile motion, angle of projection problem.)
     
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