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Find the range of the golfball and the maximum height reached

  1. Apr 15, 2012 #1
    1. The problem statement, all variables and given/known data

    A golf ball is hit with a velocity of 22.8m/s at an angle of 40.0° with respect to the horizontal.

    a) Find the range of the golf ball.
    b) Find the maximum height reached by the golf ball.

    2. Relevant equations

    dv=1/2*at2
    dh=Vh*Δt
    Kinetic equation d=Vi*t+ 1/2*at2
    dh=-V2*sin2θ /g
    sinθ=opp/hyp
    cosθ=adj/hyp
    Δt=-2Vsinθ / g

    3. The attempt at a solution

    a) dh=-V2*sin2θ /g= 52.24m

    b)dv=sin(40)*22.8=14.7m

    I am unsure of which equations to use in this problem but I've attempted something that I think may be right but would greatly appreciated if someone could just check my answers and correct me if necessary and explain which equations you use if applicable. Also if you could let me know if all my significant figures are being respected, I still have a hard time remembering the rules... if anyone had a trick for remembering the sig fig rules and would like to share that would be awesome! Thanks so much in advance!
     
  2. jcsd
  3. Apr 15, 2012 #2

    BruceW

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    You have done a) correctly, but b) is not correct. To work out the max height of the ball, first think about the ball's motion, at what stage in its journey will it be at its highest point?
     
  4. Apr 16, 2012 #3
    Thank you for your help! I don't understand why my part b) is wrong however. I don't know any other equation to calculate the vertical distance.
     
  5. Apr 16, 2012 #4
    for part b) would it be Δy=Vi2/-2ay

    Which would turn out to be Δy=26.5m?
     
  6. Apr 16, 2012 #5
    for constant acceleration kinematics in one dimension
    1. d=do + vot + 1/2at2 or
    2. vf2 = vi2 + 2ad

    For a motion in plane, resolve it to one dimension components. Eg. x/h-direction and y/v-direction.
     
  7. Apr 18, 2012 #6

    BruceW

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    Very close. Vi should be the vertical component of the initial speed, but you have used the total initial speed.
     
  8. Apr 22, 2012 #7
    for b) i tried to redo it and i used H=Visin2θ/2g and with that equation I got 10.96m. Does that answer seem correct?
     
  9. Apr 22, 2012 #8

    BruceW

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    The answer is correct. (although I got 10.95m which is slightly different). But I think you've written up the equation incorrectly. It should be Vi2 But I'm guessing you just had a miss-type?
     
  10. Apr 22, 2012 #9
    dani123, for the reason why you're answer is slightly off Bruce's see my answer to your other post on projectile motion
     
  11. Apr 22, 2012 #10

    BruceW

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    oh, I see. The difference is because I used g=9.81 and dani123 used g=9.8 This isn't a big issue. Maybe it would be better to use more significant figures in the calculation than you use in the answer, to get a more accurate answer.
     
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