Solving 2D Motion: 80m Cliff, 1330m Distance

• clope023
In summary, the shell is fired from the top of an 80 meter high cliff with a horizontal velocity of 80.6 meters per second. The shell impacts the ground 1330 meters from the base of the cliff.
clope023
[SOLVED] 2D motion

A shell is fired with a horizontal velocity in the positive x direction from the top of an 80 m high cliff. The shell strikes the ground 1330 m from the base of the cliff.

http://courses.science.fau.edu/~rjordan/quizzes/phy2043/images/3.07.gif

I think relevant equations would be x = (v0cosα0)t

y = (v0sinα0)t-1/2gt^2

vx = v0cosα0

vy = v0sinα0-gt

R = v0^2(2sinα0)/g

I think α0 = 90 degrees?

I used 1330m = v0^2(2sin90)/9.8, got v0 = 80.6m/s, one of the answer choices was 82m/s but it was wrong; the answer turned out to be 330m/s; I don't know how they do it with the info presented, any help is appreciated.

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I think you're making the problem more difficult than it is.

Based on the wording and picture of the problem, how much of initial velocity is horizontal, and how much is vertical? (Hint - you don't need trig functions).

chocokat said:
I think you're making the problem more difficult than it is.

Based on the wording and picture of the problem, how much of initial velocity is horizontal, and how much is vertical? (Hint - you don't need trig functions).

hmm, so I got the time doing what you said, treating it as just a kinematic equation with the one variable I knew, 1/2g, so

80m = 1/2gt^2, which got me t = 4s

should I plug that into one of the kinematic equations? I posted? or perhaps use one fo the regular motion along a straight line equations since I now know time?

Yes, you're exactly on the right track. Plug it into the equation:

$$\Delta x = v_0 t + \frac{1}{2} a t^2$$

where a = 0 so that whole piece goes away, and solve for v and you're all set.

BTW, I got t = 4.04s which gets you much closer to the actual answer than just t = 4s.

chocokat said:
Yes, you're exactly on the right track. Plug it into the equation:

$$\Delta x = v_0 t + \frac{1}{2} a t^2$$

where a = 0 so that whole piece goes away, and solve for v and you're all set.

BTW, I got t = 4.04s which gets you much closer to the actual answer than just t = 4s.

hmm, yeah I usually round off but maybe I should just stick with 2 decimal points after, seems to help.

I did what you said and got v_0 = 329.20m/s, close enough to the actual 330m/s that the online question wanted; thanks a lot man.

1. How do you calculate the initial velocity of an object from a given cliff height and distance?

To calculate the initial velocity (vi) of an object, we can use the formula vi = √(2gH), where g is the acceleration due to gravity (9.8 m/s²) and H is the height of the cliff. We can also use the formula vi = d/t, where d is the distance and t is the time it takes for the object to travel that distance.

2. What is the maximum height reached by the object during its motion?

The maximum height reached by the object can be calculated using the formula h = H + (vi²sin²θ)/(2g), where H is the initial height, vi is the initial velocity, θ is the angle of launch, and g is the acceleration due to gravity.

3. How long does it take for the object to reach the ground?

To calculate the time (t) it takes for the object to reach the ground, we can use the formula t = √(2H/g), where H is the initial height and g is the acceleration due to gravity. This will give us the total time for the object to reach the ground from the initial height.

4. How do you determine the horizontal and vertical components of the object's velocity?

To determine the horizontal and vertical components of the object's velocity, we can use the formulas vx = vi cosθ and vy = vi sinθ, where vi is the initial velocity and θ is the angle of launch. The horizontal component (vx) will remain constant throughout the motion, while the vertical component (vy) will change due to the acceleration due to gravity.

5. Can the same calculations be used for objects with different mass?

Yes, the same calculations can be used for objects with different mass as long as they are in a vacuum and not affected by air resistance. In this case, the mass of the object does not affect its motion, only the initial velocity and angle of launch will change the trajectory.

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