Projectile Motion of a golf ball

  • Thread starter ada15
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  • #1
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http://www.rit.edu/~jaa5898/Projectile%20Motion%20%20Results_files/image002.gif [Broken]


Question:

A golfer strikes a golf ball at an an angle of 17 degree above the horizontal. With what velocity must the ball be struck in order to reach the green, which is a horizontal distance of 250 from the golfer at the same height.


The answer should be 66m/s 17 degree above the horizontal

I know that the range is given which is 250 m and we also know the angle it means the velocity has x and y components.
I think that height dy should be zero because the ball starts its motion and comes back to the same point . so shouldn't it b zero ??

I don't know how to approach to the answer. plz give me some help.
Thanks
 
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Answers and Replies

  • #2
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Write the equations of motion on the horizontal and vertical. The time variable is common to both the equations so you can solve the system.
 
  • #3
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i know that delta t is common in projectile motion but there is only delta d (x-component ) given and an angle, nothing else.. there is no velocity so i could use it with angle and calculate the sine law and cosin law..and not even time ..

any answer
 
  • #4
Hootenanny
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But you know that the golf ball must travel 250m in the horizontal direction, i.e.

[tex]v\cos(\theta) \Delta t = 250[/tex]

You can also write an expression for the vertical plane using the kinematic equation ([itex]s = vt + \frac{1}{2}at^2[/itex]). Once you input the appropriate values you will have two equations with two unknowns....
 
  • #5
berkeman
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Thread moved from Advanced Physics to Introductory Physics.
 
  • #6
PSOA
ada15, would it best be for you to write the trajectory equation, i. e., [tex]y(x)[/tex]. As Hootenany said, just eliminate t.
 
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  • #7
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Theres an easy approact to this. Yes, we know the formula

R=ucostheta x time

If we consider time to reach time to reach maximum height ( considering
vertical components at the point where usintheta= 0m/s)

v=u+at
0+usintheta-gt
t=usintheta/g

Now we know, when the ball lands on the ground, the displacement will be equal to zero. Hence

<< Rest of exact solution deleted by berkeman >>
 
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  • #8
berkeman
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Theres an easy approact to this. Yes, we know the formula

R=ucostheta x time

If we consider time to reach time to reach maximum height ( considering
vertical components at the point where usintheta= 0m/s)

v=u+at
0+usintheta-gt
t=usintheta/g

Now we know, when the ball lands on the ground, the displacement will be equal to zero. Hence

<< Rest of exact solution deleted by berkeman >>
fffff, please do not post exact solutions to problems in the homework forums. It is against PF homework forum policy. We are here to help students to learn, not to provide worked-out answers for them. Give hints and ideas, help to check their work, etc.

You are very welcome to help out in the HW forums, just please think more in a tutorial way. "How can I best help this person to learn how to think and solve problems for themself?"
 
  • #9
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Appologies for any inconvience I may have caused.
I have some questions relating to projectile motion,

Consider the projectile of an object on the moon.
1. Now considering the atmosphere, what would be the effects of
its horizontal and vertical motion.

I suppose that the the horizontal distance decreases due to frictional forces while the vertical distance remains constant. But why does the vertical distance remain constant?

2. Neglecting atmosphere, this time with a lower gravational field strenth than in (1).

This would surely mean the vertical distance decreases, while horizontal distance remained constant.

If I'm wrong in any statement, please correct me.
Thank you
 
  • #10
OlderDan
Science Advisor
Homework Helper
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Appologies for any inconvience I may have caused.
I have some questions relating to projectile motion,

Consider the projectile of an object on the moon.
1. Now considering the atmosphere, what would be the effects of
its horizontal and vertical motion.

I suppose that the the horizontal distance decreases due to frictional forces while the vertical distance remains constant. But why does the vertical distance remain constant?

2. Neglecting atmosphere, this time with a lower gravational field strenth than in (1).

This would surely mean the vertical distance decreases, while horizontal distance remained constant.

If I'm wrong in any statement, please correct me.
Thank you
#1 does not make sense. There is no atmosphere on the moon.

For #2, decreasing the gravitational field strength reduces the acceleration. Why would that decrease vertical distances? What are you comparing it to? If you are comparing it to earth, with the same initial velocity the vertical distance should be much greater in lower gravity.
 
  • #11
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A sphere is being projected horizontally from a distance above the surface of the moon.

I'm saying in 1.
(I know its not a fact, but)
If the moon had an atmosphere with the same g(1.67 m/s^2).

The vertical distance between each 1.00s interval would increase until reaching a constant vertical velocity where the distance in each 1.00s interval would be constant. (Terminal velocity)

the horizontal distance between each 1.00s interval would decrease due to the presence of frictional forces..


2. The same experiment was repeated but on another planet which has no atmosphere where the acceleration of free fall is less than that of the moon.
(comparing this to the moon)
vertical component of velocity at any instant
v=u(0)+gt
v=gt
so as the ball falls downwards, its velocity would increase and hence the distance between each 1.00s interval would increase.

The horizontal component of velocity would remain constant.

Am I correct?
 
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