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Projectile from cliff

  • Thread starter Kmol6
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  • #1
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Homework Statement


A projectile is fired at an upward angle of 45 degrees from the top of a 165m Cliff with a speed of 175 m/s. what will be its speed when it strikes the ground below? (Use conservation of energy and neglect air resistance.)

Homework Equations


E= Ke +Pe
1/2 mv^2 + mgh = 1/2mv^2 +mgh

v^2= Vo^2 +2a(xo-x)
??

The Attempt at a Solution


I honestly don't know where to start, or if I have the right equations? A point in the right direction would be appreciated.

 

Answers and Replies

  • #2
138
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You only need to use the conservation of energy as the problem asks.

[tex]E_i=E_f[/tex]

[tex]\frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f[/tex]

The ball is launched with a speed of 175ms at an angle of 45 degrees. What does that tell you about velocity of the x and y components? What is their magnitude? Do you know which to use when calculating the kinetic energy?

The ball is initially at the top of a hill, what is its height? What does that tell you about its initial potential energy?

When the ball hits the ground, what is its height? What does that tell you about its final potential energy?
 
  • #3
15
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The velocity of the x and y components are 175 (cos 45) = 123.7 and 175(sin 45) = 123.7
The ball is initially at 165m above the cliff, but projects up further ( i don't know how high though?) at that point the potential energy is equal to the final kinetic energy when the ball hits the ground and at that point the balls potential energy is 0.
but I'm still confused as to what I put into the equation? Is it just a 1 step or 2 and if so how do I figure out the height of the projectile using energy?
 
  • #4
gneill
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Hint: Kinetic energy is a scalar value, not a vector. So the launch angle is immaterial. What matters for KE is the speed of the mass. So you can determine the initial KE of the projectile with the launch speed alone. You can leave the mass as a symbolic value "m" for now, expecting that it will cancel out later.
 
  • #5
138
43
The velocity of the x and y components are 175 (cos 45) = 123.7 and 175(sin 45) = 123.7
The ball is initially at 165m above the cliff, but projects up further ( i don't know how high though?) at that point the potential energy is equal to the final kinetic energy when the ball hits the ground and at that point the balls potential energy is 0.
but I'm still confused as to what I put into the equation? Is it just a 1 step or 2 and if so how do I figure out the height of the projectile using energy?
What is the magnitude of the velocity? Does the launch angle matter?

That is not correct and this is an important point to make. At the peak of its trajectory, the potential energy is not equal to the final kinetic energy. For that to be true, the ball's kinetic energy would have to be zero, but if you remember what the trajectory of a projectile looks like, it's a parabola, not a vertical line, meaning the ball is still moving forward, so its total kinetic energy can't be zero. At its highest point, its vertical velocity is zero (such that it won't go up any higher), but it's horizontal velocity is still the same, so the ball does indeed have kinetic energy at that point, though less kinetic energy than it initially had.

What the conservation of energy tells is that the sum of the kinetic and potential energy is going to be constant. That means it doesn't matter which point of flight we choose to measure them. The value of their sum is always going to be the same. To make our lives easier, we can find the sum at the time when it is initially launched because we can easily find both the kinetic and potential energy at that time. That sum must then be equal to its final kinetic energy.

As you said, the final potential energy is equal to zero and so the equation becomes

[tex]\frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2[/tex]

If we take its initial velocity to be its launch velocity and its initial potential energy to be the potential energy at its initial height? What values will you use?
 
  • #6
15
1
using
1/2mv^2+mgh=1/2mv^2 + 0
The mass's cancel out, so if I fill in the following:

1/2 (175)^2 + (-9.8)(165) = 1/2 v ^2 ( my only question here is should I use the initial velocity or the horizontal velocity of 175 (sin45) = 123.7 m/s ?)
13695.5 = 1/2v^2
V = 165.5 m/s

Is this correct? The book doesn't provide this answer.
 
  • #7
haruspex
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using
1/2mv^2+mgh=1/2mv^2 + 0
The mass's cancel out, so if I fill in the following:

1/2 (175)^2 + (-9.8)(165) = 1/2 v ^2 ( my only question here is should I use the initial velocity or the horizontal velocity of 175 (sin45) = 123.7 m/s ?)
13695.5 = 1/2v^2
V = 165.5 m/s

Is this correct? The book doesn't provide this answer.
Your method is fine. You are right to use magnitude of the initial velocity since it all contributes to the KE.
But you should have been surprised to find it lands with less speed than it had to begin with. (Tip: always stop to check whether an answer makes sense.)
What do you think you might have wrong?
 
  • #8
15
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I've been questioning the height this whole time, as it is fired at a 45 degree angle the y component isn't just the height of the cliff, but thats the logic I'm going through. and if the height is in fact higher then 165m , how do I calculate it when I only have velocity? Also, I did notice that the final velocity was smaller and again it didn't make sense but I don't know where to go from here?
 
  • #9
haruspex
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, I did notice that the final velocity was smaller and again it didn't make sense
1/2 (175)^2 + (-9.8)(165)
Does falling the 165m increase the KE or reduce it?
 
  • #10
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As it is falling the KE is increased and PE is decreased
 
  • #11
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OK so the 165 should be -?

16938.5=1/2v^2

V= 184.05 m/s
 
  • #12
haruspex
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As it is falling the KE is increased and PE is decreased
Right. Is that what your equation says? Is 1/2 (175)^2 + (-9.8)(165) more or less than 1/2 (175)^2?
 
  • #13
haruspex
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OK so the 165 should be -?

16938.5=1/2v^2

V= 184.05 m/s
Yes.
 
  • #14
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Oh boy, thank you!!!
 

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