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Horizontal Projectile Motion

  1. Sep 18, 2011 #1
    1. The problem statement, all variables and given/known data

    A tennis ball is thrown horizontally from an elevation of 13.03m above the ground with a speed of 24.0m/s.

    a)where is the ball after 1.50 seconds (both horizontally and vertically)?

    b) if the ball is still in the air, how long before it hits the ground?

    c) if the ball is still in the air, where will it be with respect to the starting point once it lands?

    I'd really appreciate it if you'd be able to help me as soon as possible. Thanks!


    3. The attempt at a solution

    for the first part, I use the displacement equation and found the horizontal displacement to be 36m and the vertical displacement to be 2.005m. For part b and c, I'm not sure how to start the question.
     
  2. jcsd
  3. Sep 18, 2011 #2
    For simplicity, lets make a table of a given data. :smile:
    339htug.png

    H-Horizontal Motion
    V-Vertical Motion
    sH-Displacement in horizontal direction
    sV-Displacement in vertical direction
    u-initial velocity
    t-time

    Okay, for the first part, how you get your displacement to be 2.005m for the vertical motion? :confused:
    Show some steps. :smile:
     
  4. Sep 18, 2011 #3
    I used the equation d=Vy(t) + 0.5(ay)(t^2)
    =(0)(1.5) + (0.5) (-9.8)(1.5^2)
    =2.005m
     
  5. Sep 18, 2011 #4

    rl.bhat

    User Avatar
    Homework Helper

    Check the calculations.
    d = (0.5)(9.8)(2.25) = ..?
     
  6. Sep 18, 2011 #5
    you get -11.025m from the displacement equation and add 13.03m because its that much higher than the ground. I have confirmed that this answer is correct but am unsure how to do part b and c.
     
  7. Sep 18, 2011 #6
    For part b, you can do this:-
    We have to calculate the time when it reaches the ground.
    For that you can have 13.03m in the displacement equation and find the time.

    For part c:-
    When you get the time for vertical motion from part b, it would be same for horizontal motion.

    Try solving. :smile:
     
  8. Sep 18, 2011 #7
    if the ball is 36m horizontal and 2.005m vertical at 1.5 seconds, for part b if I set d=13.03 and use v=24m/s, then t=0.54s. This doesn't make sense because at 1.5 seconds its still in the air so how can it hit the ground at 0.54s?
     
  9. Sep 18, 2011 #8
    2.005m is the height above the ground at 1.5 seconds. :smile:
    You used initial velocity as 24m/s for vertical motion, which is wrong. The initial velocity for vertical motion is 0m/s. Check the table above. :)
     
  10. Sep 18, 2011 #9
    okay so for part b, I put d=13.03 v=0, a=9.8 and used the displacement equation to find time which gave me 1.63s...but this was wrong. But for part c, when I used 1.63s, it gave me the write displacement answer. I am unsure as to what I did wrong with part b.
     
  11. Sep 18, 2011 #10
    What's the answer then, in case if you have the answers? :confused:
     
  12. Sep 18, 2011 #11
    I don't have the answers- it is an online question so when I get it wrong, it just says wrong. would you have any other suggestions??
     
  13. Sep 18, 2011 #12
    Are there any options, if so please post them here. :)
     
  14. Sep 18, 2011 #13
    no there aren't any options! sorry, but would you be able to help me with the following question?:

    A cannonball is catapulted toward a castle. The cannonball's velocity when it leaves the catapult is 32 m/s at an angle of 50° with respect to the horizontal and the cannonball is 7.0 m above the ground at this time.

    a) Assuming the cannonball makes it over the castle walls and lands back down on the ground, at what horizontal distance from its release point will it land?

    For this question, I used the equation for Range and got 102.9m but it was incorrect.

    b)What is the y-component of the cannonball's velocity just before it lands? The y-axis points up.

    For this question, I did Vy=32sin50=24.5 (but this was also incorrect). Any help will be appreciated!
     
  15. Sep 18, 2011 #14

    rl.bhat

    User Avatar
    Homework Helper

    1.63s - 1.5s = .? is the required answer. Check it.
    For the second problem, try the following formula.
    -y = x*tanθ - 0.5*g*x^2/(v^2*cos^2θ) and solve for x.
    For part b, the ball is landing bellow the point of projection. Find time of flight and use
    -vy = -v*sinθ - gt. to get vy.
     
    Last edited: Sep 18, 2011
  16. Sep 18, 2011 #15
    yes that was the correct answer. thank you!
     
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