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

[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.

 

[chocokat]

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).

 

[clope023]

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?

 

[chocokat]

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.

 

[clope023]

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.