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Projectile Motion(Fun. Fun) where object is thrown from ground level

  1. Oct 18, 2009 #1
    Ok basically I've tried loads of methods of working it out and I'd just like to see how someone else would do it because all the answers I get don't seem right and I am not confident about them. A bit of explaining of why things that may seem odd are wouldn't be minded elther.
    I've tried working out the speeds on both planes using 15Sin(30 and 15cos(30 but when I try and apply the answer to my motion equations S = ut + (0.5)(A)(t squared) and V(Squared) = U(Squared) + 2as, It doesn't seem to work out elther way. The way I did it in the end was using V = u + at working out U with sin and assuming a = -9.81 and V = 0.

    Here's the question:
    A ball is thrown from the ground with a velocity of 15ms at an angle of 30 to the horiziontal.
    Calculate
    A . It's time of flight(The time between the point at which it leaves the thrower's hand when it hits the ground) assuming that the ground is level.
    B. It's range.
    C. It's maximum height.

    -Shoot.
     
    Last edited: Oct 18, 2009
  2. jcsd
  3. Oct 18, 2009 #2

    Mark44

    Staff: Mentor

    You need to decompose the initial velocity vector into its horizontal and vertical components. The horizontal component has a magnitude of 15cos(30deg.) = 15(sqrt(3)/2) m/sec. The vertical component's magnitude is 15/2 m/sec.

    Look at the kinematic equations for the motion vertically and for its motion horizontally. If the ball were thrown upward at a speed of 15/2 = 7.5 m/sec, the only force acting on it is the force of gravity, so you should be able to figure out when (the value of t) the ball comes back to the ground. That will be the answer to A. You can then substitute this value of t in your equation for horizontal position, and that will give you the answer to B. For C, the maximum height occurs midway through the ball's flight.
     
  4. Oct 19, 2009 #3
    Ok so if I try 15Sin(30 and work out u is 7.5 m/s and then put this into V = u + at where V = 0, And work out T that way as the ball's time is how to be effected how high it goes and then falls. I get 0.765(Sec) with this - Is this correct?

    Oh and my working out is that accleration is minus as the ball is being thrown upwards so A = G = - 9.81 because an object being thrown up should deacclerate.

    So input that into V = u + at gives me 0 = 7.5 + (-9.81)(t) and then changing the subject so -9.81t is on the otherside and so turned into a plus, And then diviiding 7.5 by 9.81.

    - That's my answer to A so far, I want to check that is correct before I move on.
     
  5. Oct 20, 2009 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    It would help if you would tell us what "u" means! Is it the x or y component of velocity? And why are you "trying" 15 cos(30) and 15 sin(30)? What are you trying to use them for?
     
  6. Oct 20, 2009 #5
    Oh of course sorry - I'm too used to just doing this in my exercise book as notes.
    U is initial velocity of the traveling object and I am working it out in the Y axis.
    I'm using 15Sin(30 because I need to use that to work out the time, The time is going to be the time it traveled in the parabola instead of the straight line you'd get on the X axis.
     
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