Projectile launched over cliff

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In summary, the problem involves a projectile being launched from a cliff with a speed of 50.0 m/s and an angle of 40.0o above the horizontal. The projectile lands on a flat valley floor at the base of the cliff, with a height of 75.0 m. Using equations for projectile motion and assuming g = 10.0 m/s2 and no aerodynamic effects, the time it takes for the projectile to reach the valley floor can be calculated by breaking the initial velocity into its horizontal and vertical components using trigonometry and setting the upward direction as positive. The equation used should result in both positive and negative values for time, with the positive value being the desired answer.
  • #1
Jordash
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Homework Statement



A projectile is launched with a speed of 50.0 m/s and an angle of 40.0o above the horizontal, from the top of a 75.0 m high cliff onto a flat valley floor at the base of the cliff. Assume that g = 10.0 m/s2 and ignore aerodynamic effects.

How long does it take for the projectile to reach the valley floor?



Homework Equations





The Attempt at a Solution



Not sure on this one
 
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  • #2
Hi Jordash, well this problem again is utilities the equations that I gave you in your thread on a https://www.physicsforums.com/showthread.php?t=286763":
Galadirith said:
[tex]v_f = v_i + at[/tex]
[tex]s = \frac{(v_f + v_i)}{2}t[/tex]
[tex]s = v_it + \frac{1}{2}at^2[/tex]
[tex]s = v_ft - \frac{1}{2}at^2[/tex]
[tex]v_f^2 = v_i^2 + 2as[/tex]

Now you may not be familiar with the fact that if you give something a velocity at an angle, you can break it up into its horizontal and vertical components using trigonometry. Once you have done that you consider the vertical and horizontal motion of the particle seperatly.

You should also define which direction you want to be positive, for example is down negative which would would make the acceleration due to gravity negative, it doesn't matter which way you chose, as long as you are consistent :D

With that try and have an attempt at the question, then post back here with as far as you can get yourself :D
 
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  • #3
Ok, the angle still throws me a little bit off, because won't the displacement (s) be more than 75 meters? Because it is shot up 40 degrees into the air? How do you take that into consideration?

Thanks
 
  • #4
Oh k sure. so let's consider it like this, let's set the zero point of vertical displacement at the top of our cliff. Let us also say that the upward direction is positive. Now assuming you are comfortable with breaking the initial velocity into its components using trigonometry, you will find that the initial vertical component of velocity to be positive (pointing in the upward direction).

Now we have set the zero point to the top of the cliff, what this means is that the final displacement (which is a vector quantity so must have a direction, in this case up or down) is negative 75m, can you see that. It actually doesn't matter about the hight that it goes up to, the equation does need to know that :D.

Now (this will give you a little bit of a hint to the correct equation to use) if you chose the correct equation you will actually end up with a positive and negative value for time. obviously it should seem logical that you want the positive value for time, and if you would like I will explain where the other value comes from.

So try setting up you equation now, paying special attention to the direction in which you values are pointing (eg gravity is acting down so will that be positive or negative in the equation)
 
  • #5
Now assuming you are comfortable with breaking the initial velocity into its components using trigonometry, you will find that the initial vertical component of velocity to be positive (pointing in the upward direction).

Sorry I'm still a bit confused I first tried solving for Velocity Final LaTeX Code: v_f = v_i + at and did vf=0+2(-10m/s)(-75m) which gave 1500 and then I tried to solve for time using LaTeX Code: s = \\frac{(v_f + v_i)}{2}t and that came out with .1 second but I don't think I'm doing it right. :(
 

Related to Projectile launched over cliff

1. What is a projectile launched over a cliff?

A projectile launched over a cliff is an object that is thrown or propelled off a cliff, typically with an initial horizontal velocity. It is affected by the force of gravity and follows a curved path called a trajectory.

2. How does the height of the cliff affect the projectile's motion?

The height of the cliff affects the projectile's motion by increasing the time it takes for the projectile to reach the ground. This is because the higher the cliff, the longer the projectile is in the air and the longer it is affected by the force of gravity.

3. What factors influence the range of the projectile?

The range of a projectile launched over a cliff is influenced by the initial velocity, the angle at which it is launched, and the height of the cliff. A higher initial velocity and a lower launch angle will result in a longer range, while a higher cliff height will decrease the range.

4. Can the projectile's motion be accurately predicted?

Yes, the projectile's motion can be accurately predicted using mathematical equations such as the equations of motion and the range formula. These equations take into account the initial conditions and the forces acting on the projectile.

5. What is the significance of studying projectiles launched over cliffs?

Studying projectiles launched over cliffs is important in understanding the laws of motion and gravity. It also has practical applications in fields such as engineering and ballistics, where the motion of projectiles is crucial for accurate calculations and designs.

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