Need clarification again

  • Thread starter frankfjf
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n Fig. 4-33, a stone is projected at a cliff of height h with an initial speed of 40.0 m/s directed at an angle 0 = 56.0° above the horizontal. The stone strikes at A, 5.10 s after launching. Find (a) the height h of the cliff, (b) the speed of the stone just before impact at A, and (c) the maximum height H reached above the ground.

Part c is basically asking for the trajectory, right?
 

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  • #2
Doc Al
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frankfjf said:
Part c is basically asking for the trajectory, right?
In a sense, yes, but only one aspect of it. How would you go about finding the maximum height?
 
  • #3
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To be honest I'm not sure.
 
  • #4
Doc Al
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Here's a hint: What's the vertical component of the velocity when the stone reaches its highest point?
 
  • #5
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That's just it, I have no idea what to plug in for determining the velocity at the highest point.

I think that the formula involved would be the vertical motion equation for velocity, Vy = Vi*sin(angle) - gt.

Well actually, if I took half of the time, wouldn't it be at its highest point then?
 
  • #6
Doc Al
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frankfjf said:
That's just it, I have no idea what to plug in for determining the velocity at the highest point.
Think about it. Imagine throwing a ball straight up in the air. Initially it moves up (velocity = +); after reaching the highest point, it starts moving down (velocity = -). So what can you say about the speed at the highest point?

I think that the formula involved would be the vertical motion equation for velocity, Vy = Vi*sin(angle) - gt.
Good.

Well actually, if I took half of the time, wouldn't it be at its highest point then?
Only if the initial and final positions were at the same height, which is not true in this problem.
 
  • #7
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Doc Al said:
Think about it. Imagine throwing a ball straight up in the air. Initially it moves up (velocity = +); after reaching the highest point, it starts moving down (velocity = -). So what can you say about the speed at the highest point?
Well, at the highest point the speed would be zero, since it's going to start falling immediately after hitting the highest point.
 
  • #8
Doc Al
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Exactly right. Now make use of that fact.
 
  • #9
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But now I'm confused as to just where to plugin the zero.

If the formula Vy = Vi*sin(angle) - gt is used, does the zero replace Vy or Vi?
 
  • #10
Doc Al
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What do you think? (Does the initial velocity change?)
 
  • #11
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No, but if I put in zero for Vy, I'm left without any variables.
 
  • #12
Doc Al
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You can solve for the time; that's what you need.
 
  • #13
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Ahhh I got it now, thanks!
 

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