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How long is the ball airborne

  1. Sep 25, 2005 #1
    i have a test tommorw and i need answeres to the folowing questions if u can help i would be very thankful
    1.
    a bout can travel at 1.85 m/s in still water at wat angle must it go in order to go stright across a 110 m river flowing at 1.20m/s
    b)
    if it goes stight across detrime the time it would take
    c)
    if it goes stright across how far down streem would it end up

    Q2

    a child throws a baseball onto the roof a house then catches it 1.0 m from the ground the roof has an angle of 33 and from the ground to the tip of the roff the distance is 6.2 m is the ball fals at a speed of 3.2 m/s
    a)
    how long is the ball airborne
    b)
    wat is the horizantal distance form the glove to the edge of the roof
    c)
    wat is the velocity of the ball just before it hits the glove

    plz i need help with this i have a test and i need to know it i really need ur help
     
  2. jcsd
  3. Sep 25, 2005 #2
    I'll address problem one first.

    Draw yourself a picture of what is going on. You are crossing a river that is flowing perpendicular to your motion. So what kind of triangle is setup here? You guessed it. A right triangle. So when you're drawing your triangle make sure that every component is of the same unit. Don't use meters and meters per second in the same triangle.

    To compensate for the flow of the river, you will travel at some angle, [itex]\theta[/itex], and the result will be moving directly across. So which parts of the right triangle do you have? What trig inverses can you use to find the angle in which you are moving?
     
  4. Sep 25, 2005 #3
    well i sued sin and got an angle of 126 but that would go with the current do i flip it and get the other side or wat
     
  5. Sep 25, 2005 #4

    EnumaElish

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    For #1 my approach would be to make a drawing and see what the drift is if the boat starts with a right angle to the direction of the waterflow. I'd then calculate how much the drift is reduced if the boat starts with a 1-degree angle toward the flow. Finally I'd ask how many degrees are needed to reduce the drift to zero.
     
  6. Sep 25, 2005 #5
    well would it no be easier to do it with the sin
     
  7. Sep 25, 2005 #6
    any way if u take 90 out of 126 u would get 36 degrees against the drift
     
  8. Sep 25, 2005 #7

    EnumaElish

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    can you explain how you got 126?
     
  9. Sep 25, 2005 #8
    252 sin 30
     
  10. Sep 25, 2005 #9
    ohhh i was looking at the wrong q
     
  11. Sep 25, 2005 #10
    i used the sin law to fin thada and rearanged it
    i got 14.17
     
  12. Sep 25, 2005 #11
    first i used the pathagrian therom to find the hypotinous then used that side and 90 angle andthe side oppsite to thada and thada to figer out thada
     
  13. Sep 25, 2005 #12

    EnumaElish

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    And how did you use the Pythagoran Theorem? Can you write it for my benefit?
     
  14. Sep 25, 2005 #13
    a^2=b^2+c^2
    wich is
    a^2= 1.85^2 + 1.20^2
    actually my answere is wrong it should be for the hypotinous 2.205...
     
  15. Sep 25, 2005 #14

    EnumaElish

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    So how did you figure out theta (the angle) from this?
     
  16. Sep 25, 2005 #15
    then u se the sin law
    ie. sin thada divided by 1.20 = sin 90 over 2.2 then u cross multiply to get sin thada alone then u use the invere of sin to get theda
     
  17. Sep 26, 2005 #16

    EnumaElish

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    So what's the angle you have found? You said there was a mistake, have you corrected it? What is the angle (theta) you are finding after you correct the mistake?
     
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