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Homework Help: Another question about Projectiles

  1. Oct 6, 2012 #1
    A tennis ball is thrown straight up at 10 m s -1 from a car travelling at 30 m s -1

    a)Determine the time of flight for the ball
    b)How far will the car have travelled before the ball lands in the car?
    c)Why in practice would your answer to a) not happen

    S = ut
    v = u + at

    a) At first I thought it was straight forward but I think I may be confusing myself here:
    My view on this is that the question is asking determine how long the ball is in the air for so ,

    Initial vertical velocity = 10
    a = -9.8
    Final vertical velocity = 0

    Using v = u + at
    0 = 10 + -9.8xt
    9.8t = 10
    t = 1.02 s

    Then I was thinking is this the time it takes for the ball to reach maximum height (Since v=0 at max height) or is this the time it takes to go up and fall back down ? Who can point me in the right direction here.

    b) Straight forward , S = ut
    S = 30x1.02
    S = 30.6m

    c) Unsure , thought about it and was thinking it could be to do with the fact the car is moving at 30m s -1 ?

  2. jcsd
  3. Oct 6, 2012 #2


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    Yes, for precisely the reason you gave.
    You'll need to use the correct time here.
    No, if the car moves forward at 30 m/s while the ball also moves horizontally at 30 m/s, then the ball will "land in the car". They are asking why this would not really happen in practice.
  4. Oct 6, 2012 #3
    Ok so when v=0 (At max height) that is the time taken to reach max height and is different from when v=0 (Ball on ground) But I've never thought about that all this time and my answers to other questions involving v=0 was right (when an object stops) So you're saying that what I wrote is the time taken for the ball to reach max height ? If that's true then I must multiply my answer by 2 since at max height half the journey has already been completed.

    time to reach max height (v=0) = 1.02 s
    time to go up and fall back down (v=0) 1.02 x 2 = 2.04s

    How far has the car travelled since then would be
    S = 30 x 2.03
    S = 61.2m

    As for c) I am still unsure what it means by why it won't happen.
  5. Oct 6, 2012 #4


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    What influence on the projectile do we always neglect in projectile motion problems?
  6. Oct 6, 2012 #5
    Air resistance
  7. Oct 6, 2012 #6


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    Yes. 30 m/s is more than 65 mph. Imagine the air resistance on your hand if you stick it out the window of a car moving that fast!
  8. Oct 6, 2012 #7
    I am still unsure about a) and b) are my new answers correct? Also how would you know that if t was 1.02 seconds it applies for the time taken to reach maximum height and not the time taken to go up and down without thinking about it?
  9. Oct 6, 2012 #8
    Should be time taken to go up and down! o_O
    Time of flight, even if its going down, its considered a flight right? :tongue:
    So your answer for a) should be 1.02x2=2.04s (time to reach the top and drops down again is equal!)
    Actually you can find the time taken directly from one single equation.
    s=ut+0.5at^2 <-----this equation :tongue:

    for b) your equation is correct, just use the new time you calculated from a)
    Last edited: Oct 6, 2012
  10. Oct 6, 2012 #9
    Eh sorry just noticed you have found the correct answer :tongue: careless me.
    About when to find what...
    at the maximum height, velocity is equal to zero, as in your equation.
    when the ball returns to the surface, displacement is equal to zero and velocity is negative to the same magnitude of initial velocity.
    Last edited: Oct 6, 2012
  11. Oct 6, 2012 #10


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    As you noted earlier, you found the 1.02 s by setting the final y-component of velocity = 0. There is only one place at which that occurs - namely, at maximum height. So, that's how you know that you found the time at maximum height.
  12. Oct 6, 2012 #11


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    If you are not considering air resistance, then the "car travelling at 30 m s-1" is irrelevant! This problem is exactly the same as if we threw the ball up while standing still. Of course "in practice" we can't ignore air resistance.
  13. Oct 6, 2012 #12
    Haha in practice you will get fined for throwing a ball out of your car :P
  14. Oct 6, 2012 #13
    Ok I've taken in as much information from what everyone has said and I've come to this realisation :

    Is there really 2 times where v=0? At maximum height realistically yes that is true but I was assuming that when the ball returned to ground v=0 implying that it stopped straight away which is not the case also the other formula hadn't come to mind (s = ut + (at^2)/2)
    The final velocity of the ball just before it hits the ground should be greater than 0 so the flight is complete right?

    As for the second part of this scenario , If I were to use v = u + at for maximum height and the time it takes to make its flight (go up and down) shouldn't it still be right? Unless v is not 0 when it returns to the ground. So is it safe to assume that v is not 0 when it hits the ground because that's the way I see it.

    Suppose the question changed to

    A ball is thrown up with an initial velocity of 10 m s -1 .
    a)Calculate the time it takes for the ball to reach maximum height
    b)How long would it take for the ball to hit the ground?

    I would have used v = u + at here assuming that v=0 applies to both of them.

    What I'm really trying to say is that
    v=0 can't be true for both right?

    The last part question c) I really don't get , why wouldn't it take 2.04 seconds to make it's trip , because there is air resistance which affects the velocity of the ball?
  15. Oct 6, 2012 #14
    v=0 only when its maximum height.
    v=-u when it reaches the ground! :)
    notice the negative sign I put there. This shows that the ball is travelling downwards now.
  16. Oct 6, 2012 #15
    about the c) part.

    Imagine yourself in a car.
    You wind down the window and stick your face out of it.
    What will you feel? Wind right?
    The faster the car goes, the faster is the wind :)
    This is because air resistance is related to your velocity, you are smashing your face into the air particles in the atmosphere at high speed!
  17. Oct 6, 2012 #16
    Ok I'm past the confusion with the ball , it's because of air resistance that will affect the velocity of the ball ! Thanks everyone
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