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Distance traveled and the kinetic energy of a sphere

  1. Oct 4, 2015 #1
    1. The problem statement, all variables and given/known data
    A 'hammer' thrown in athletics consists of a metal sphere, mass 7.26 kg, with a wire handle attached, the mass of which can be neglected. In a certain attempt it is thrown with an initial velocity which makes an angle of 45 degrees with the horizontal and its flight takes 4 seconds.

    Find the horizontal distance travelled and the kinetic energy of the sphere just before it strikes the ground, stating any assumptions and approximations you make in order to do so.

    Answers: 80 m, 2900 J

    2. The attempt at a solution
    s = 1/2 at2 = 1/2 * 10 ms-2 * 42 = 80 m

    KE = 1/2 mv2 = 1/2 * 7.26 * ?

    v = u +at = 0 + 10 ms-2 * 4 = 40 ms-1
    v2 = u2 +2as = 0 + 2 * 10 ms-2 * 80 m = 402 m

    KE = 5808 J, wrong. Any tips please? Also can't think of a way to use the given angle.
     
  2. jcsd
  3. Oct 4, 2015 #2
    This is an equation for the distance traveled by an object falling from rest. In this question you need to consider the horizontal and vertical components of motion separately. This is where the angle comes into play for the horizontal distance traveled. How fast is the shadow of the object moving? This is the same as the horizontal speed.

    As for the kinetic energy - how does the speed of the hammer upon impact compare to its speed when thrown? Why?
     
  4. Oct 4, 2015 #3
    1. s = ut + 1/2 at2 = 0 * 4 + 1/2 * 10 * 42 = 80 m
    2. v = u + at = 0 + 10 * 4 = 40 m s-1
    vHorizontal = 40 cos 45 = 28.26 ms-1
    KE = 1/2 mv2 = 2904 J
     
  5. Oct 4, 2015 #4
    Good.
    Edit: the 40m/s initial speed is incorrect - see post 6

    It looks like you have confused the horizontal and vertical distances. What is s? The horizontal distance? Is there an acceleration in the horizontal direction (the second term in your equation says that there is if I have interpreted you correctly)?

    Ok good. Do you understand why the velocity is the same at impact as it is when launched/thrown?
     
    Last edited: Oct 4, 2015
  6. Oct 4, 2015 #5
    h = height (vertical)
    s = distance (horizontal)

    Well the horizontal movement is v cos 45 and vertical movement is v sin 45, where v = velocity. If v = u + at = 40 m/s then both hor. and ver. movements are 28.28 m/s (40 cos 45 and 40 sin 45). But I don't know how to get the correct distance from that. s = vt = 28.28 m/s * 4 s = 113 m.

    Hm I guess because there is no resistance and that's why it should be the same at impact as at start.
     
  7. Oct 4, 2015 #6
    Ah. I should have paid closer attention to one of your previous responses and the question - the initial speed of the object is NOT 40 m/s, nor is the horizontal speed 28m/s. Sorry about that. Let's go back to the beginning.

    So the object takes 4 seconds to hit the ground from being launched. If we approximate the acceleration as 10m/s/s downward then how fast must the object be traveling in the vertical direction initially? You should also be able to reason the horizontal speed from this information (you know the vertical leg of the triangle and the angle, so the horizontal leg can be found). Reasoned correctly you should see why s = v(horizontal)*t

    Indeed. The horizontal speed does not change and the vertical speed is symmetric (same speed at same height) so the impact speed must be equal as launch speed for equal heights.
     
  8. Oct 4, 2015 #7
    t = 4 s
    a = - 10 m/s2

    v = u + at
    -u = at
    -u = -10 * 4
    u = 40 m/s, the velocity at t = 0

    40 sin 45 = 28.28 m/s, vertical component of the initial velocity which is equal to the horizontal.

    s = vt = 28.28 * 4 = 113 m

    If 80 = v * 4 then v = 20 m/s, no idea how it is derived.
     
  9. Oct 4, 2015 #8
    Not quite. The 4 seconds is the time of flight. Not the time to have the vertical speed reach zero. This takes half the time.

    If the object is launched with some (unknown) initial velocity at any angle and it spends 4 seconds in the air then the vertical component must be 20 m/s because it has to slow down and come to rest (which takes 2 seconds) and then speed up and hit the ground (another 2 seconds). So if the vertical component is 20m/s then you can also reason the horizontal component is 20m/s. If both components are 20m/s then what is the initial speed (use Pythagorean theorem)? You can then find the kinetic energy upon impact.
     
  10. Oct 5, 2015 #9
    m = 7.26 kg
    t = 4 s
    a = -10 m/s2

    1. Hor. distance?
    s = vt
    v = ?
    a = (v - u) / t
    v = 0, then: -10 = - u / 2
    u = 20

    s = 20 * 4 = 80 m

    2. KE sphere?
    KE = 1/2 mv2 = 1/2 * 7.26 * (20 cos 45)2 = 726 J. Wrong.
     
  11. Oct 5, 2015 #10
    Or:
    1. Considering the vertical motion:
    u = X sin 45
    a = -10 ms^-2
    t = 2 s (max height is at half travel time)
    v = 0 at max height

    a = (v-u) / t

    -10 = (0-xsin45) / 2
    -20 = - x sin 45
    x = 20 / sin 45 = 28.28

    So the projected velocity is 28.28.

    Considering hor. motion:
    s = vt = 28.28 cos 45 * 4 = 80 m

    2. KE = PE
    KE = 1/2 mv^2
    PE = mgh

    h: v2 = u2 + 2as
    0 = 202 + 2 (-10) s
    400 -20s = 0
    s = 20 m

    PE = 7.26 * 10 * 20 = 1452 J for half the travel. 1452 * 2 = 2904 J.

    KE = 1/2 * 7.26 * 28.282 = 2904 J.

    Did I get it right?
     
  12. Oct 5, 2015 #11
    Yes.

    Great.

    Absolutely.

    Nice work.
     
  13. Oct 6, 2015 #12
    Thank you, finally got it right :).
     
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