Projectiles - golf ball

  • Thread starter Ihsahn
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  • #1
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Ok, having studied projectiles over the past few weeks, reaching the "Equation for the path of a projectile stage," I've been presented with the following question. As far as I can see, there isn't enough info to answer the question (unless, of course, its one of those simultaneous equation ones, or I've missed something obvious)

A golf ball is struck from a point O, leaving O with a speed of 21 root 2 m/s. The ball lands, without previously bouncing, at a point A, which is 10 metres below the horizontal plane through O. Find the speed of the ball at the instant it lands at A

In the question, g (Gravity) is taken to be equal to 9.8m/s^2...

Any help would be much appreciated!
 
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Answers and Replies

  • #2
Doc Al
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There are several ways to solve this. If you've covered conservation of energy, use it.

Assuming you haven't, then considered the x and y components of the velocity separately. And realize that [itex]V^2=V_x^2+V_y^2[/itex].
 
  • #3


Originally posted by Ihsahn
A golf ball is struck from a point O, leaving O with a speed of 21 root 2 m/s.

Excuse me, but I'm not familiar with the expression 21 root 2. What is it?
 
  • #4
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Originally posted by Julian Solos
Excuse me, but I'm not familiar with the expression 21 root 2. What is it?

Sorry if that was unclear, I meant 21 multiplied by the square root of 2.

Doc Al, I don't quite follow... we haven't covered conservation of energy... and I'm unsure how to obtain the x and y components given that we have no angle of projection? Forgive me if its something obvious or simple.

Thanks.
 
  • #5
Doc Al
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Originally posted by Ihsahn
Doc Al, I don't quite follow... we haven't covered conservation of energy... and I'm unsure how to obtain the x and y components given that we have no angle of projection?
You don't have to know the angle. Call the components [itex]V_x[/itex] and [itex]V_y[/itex]. Now write down the final values of those velocity components. Try it and see what you get. (Hint: Use [itex]V_f^2=V_i^2+2as[/itex].)
 
  • #6
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Again, sorry if this sounds stupid;

We've always evaluated (for example) [itex]V_x[/itex] as [itex]V cos[/itex] a and [itex]V_y[/itex] as [itex]V sin[/itex] a ??

The only thing I can think of from your suggestion is the use of trig identities? Do they play a part?

I'm obviously missing something...
 
  • #7
Doc Al
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No trig needed. Do you agree that [itex]V^2=V_x^2+V_y^2[/itex]?

So do the following...

1) Find [itex]V_{y,f}^2[/itex] in terms of [itex]V_{y,i}[/itex], [itex]a[/itex], and [itex]\Delta y[/itex] using the kinematic equation given earlier.

2) Realize that [itex]V_x^2[/itex] is a constant.

3) Add these to get [itex]V_f^2[/itex].
 
  • #8
A solution (I think )follows. If you don't want to see it, ignore this post.

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Denote the initial velocity of the golf ball [tex]V_{0}[/tex] and the final velocity [tex]V[/tex].


We have a relation

[tex] V_{0}^2 = V_{x0}^2 + V_{y0}^2[/tex].


Substituting [tex] V_{0} = 21 \sqrt {2}[/tex],

we have

[tex] 882 = V_{x0}^2 + V_{y0}^2[/tex].

Therefore,

[tex] V_{y0}^2 = 882 - V_{x0}^2[/tex].


Ignoring the effect of friction, the [tex]x[/tex] component of the speed of the golf ball remains unchanged throughout its flight.

Thus,

[tex]V_{x0} = V_{x}[/tex].


The [tex]y[/tex] component of the final speed of the golf ball upon landing is obtained from the following relation:

[tex]V_{y}^2 = V_{y0}^2 + 2g (10 \mbox{m})[/tex]
[tex] = (882 - V_{x0}^2) + 196[/tex]
[tex] = 1078 - V_{x0}^2[/tex]
[tex] = 1078 - V_{x}^2[/tex].



[tex]V^2 = V_{x}^2 + V_{y}^2[/tex]
[tex] = V_{x}^2 + (1078 - V_{x}^2)[/tex]
[tex] = 1078[/tex].


[tex]V = \sqrt {1078}[/tex]
[tex] = 32.83 \ \mbox{m/s}[/tex]
 
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