Angle that the stone is thrown off of the cliff?

  • Thread starter balling12
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  • #1
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Homework Statement


A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 16.0 m/s. The cliff is h = 62.0 m above a flat horizontal beach.

How long after being released does the stone strike the beach below the cliff?

With what speed and angle of impact does the stone land?

How do I solve without knowing the angle that the stone is thrown off of the cliff?
 

Answers and Replies

  • #2
Cyosis
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balling12 said:
A student stands at the edge of a cliff and throws a stone horizontally over the edge....
The angle is given by the bolded part.
 
  • #3
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so the angle is 0 or 180?
 
  • #4
Cyosis
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Technically zero, but what do you know about the components if an object is thrown horizontally? You don't need to do any trigonometry.
 
  • #5
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I tried working it without using trig but I got the wrong answers.
 
  • #6
Cyosis
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Show your steps so we can have a look at it.
 
  • #7
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Ok so you said I didnt need to use trig so to get the time i simply divided 62 by 9.8 and got 6.3 seconds.
 
  • #8
dx
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Ok so you said I didnt need to use trig so to get the time i simply divided 62 by 9.8 and got 6.3 seconds.
And why do you think that would give you the time?
 
  • #9
Cyosis
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That is not right, 62 meters divided by 9.8 m/s^2 gives 6.3 s^2. Do you know of any kinematic equations that give you displacement as a function of time for a uniformly accelerating object?
 
  • #10
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Because i figured if the height is 62m and the gravitational constant is 9.8m/s dividing would give me the time of the fall.
 
  • #11
Cyosis
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Well if you had just looked at the dimensions you would have figured it was wrong. As I said your calculation yields 6.3 seconds squared, which is not the unit of time.

Edit: the gravitational constant is 9.8 m/s^2 not 9.8m/s.
 
  • #12
dx
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So you simply guessed? 9.8 m/s² is the acceleration due to gravity. Do you recognize this equation: s = ut + ½at² ?
 
  • #13
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So you simply guessed? 9.8 m/s² is the acceleration due to gravity. Do you recognize this equation: s = ut + ½at² ?
yes i recognize that equation
 
  • #14
Cyosis
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What are s, u and a in this problem?
 
  • #15
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s=62 u=? a=-9.8m/s
I tried using this equation at first but ran into the same problem because i was unsure of what u is.
 
  • #16
dx
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That's the equation to use here. This is what it means: if a particle startes out with speed u (in some direction), and has an acceleration a(in that direction), then the distance travelled by the particle after time t (in that direction) is s = ut + ½at².
 
  • #17
Cyosis
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You should know what u is after reading dx's post. One thing though if you set s=62 then a needs to be positive or if you take a to be negative then s=-62. Also I generally use this notation:

[itex]
s=\frac{1}{2}at^2+v_0t+s_0
[/itex]

The v_0 symbol is slightly more suggestive than the u symbol.
 
  • #18
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so would u=16?
 
  • #19
Cyosis
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In which direction is gravity accelerating the stone? In which direction is the stone moving with 16m/s?
 
  • #20
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actually horizontally
 
  • #21
Cyosis
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Gravity is accelerating the stone horizontally? I think not.
 
  • #22
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well in the problem the stone is thrown horizonally at the speed of 16 m/s off the cliff. i understand that it is being acted on by gravity at the same time
 
  • #23
Cyosis
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Yes, but you do not appear to realize the direction of gravity. You have an equation [itex]s=62+v_0t-\frac{1}{2}gt^2[/itex]. This is what you have done correctly so far. In which direction is g acting upon the stone? In which direction does v_0 point?
 
  • #24
dx
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The equation applies to each direction. You can't mix them up. What we're interested in is the vertical direction, the direction in which the ball falls due to gravity. The initial speed in the vertical direction is what? (16 m/s is the speed in the horizontal direction)
 
  • #25
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i understand that gravity is pulling the stone down or in the negative direction.
 

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