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Find the range of the golfball and the maximum height reached

  • Thread starter dani123
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  • #1
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Homework Statement



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.

Homework 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

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!
 

Answers and Replies

  • #2
BruceW
Homework Helper
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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?
 
  • #3
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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.
 
  • #4
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for part b) would it be Δy=Vi2/-2ay

Which would turn out to be Δy=26.5m?
 
  • #5
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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.
 
  • #6
BruceW
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for part b) would it be Δy=Vi2/-2ay

Which would turn out to be Δy=26.5m?
Very close. Vi should be the vertical component of the initial speed, but you have used the total initial speed.
 
  • #7
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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?
 
  • #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?
 
  • #9
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dani123, for the reason why you're answer is slightly off Bruce's see my answer to your other post on projectile motion
 
  • #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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