Acceleration #4

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Now we get to find where something lands after being launched, but first, there’re a couple new things to learn.

s_y=s₀+v_yt+(1/2)at²

That works for the up-and-down direction. You can get a time from that and use it to find the distance in the horizontal direction.

s_x=v_xt

I’ll explain what the new variables mean. s₀ is the starting position, s_y is how far up or down it ends up, s_x is how far forward the object travels. v_y is the vertical (up-and-down) part of the object’s velocity, v_x is the horizontal (front-and-back) part of its velocity. Oh, yeah, you’ll usually get the speed (v) and angle (θ) the object gets shot at. Don’t worry, I’ll tell you how to get something useful out of that, too.

v_x=vcosθ
v_y=vsinθ

For today’s practice, we get an artillery out in Iraq (not really, but we can pretend, can’t we?). It’s aimed in the direction of terrorist camp a kilometer (1000m) away. The artillery is aimed at 30˚ above the ground and fires at 50m/s. Do you think we can hit it?

v_y=(50sin30˚)m/s
↓
0=0+(50sin30˚)m/s•t-1/2•9.8m/s²•t²
↓
0=-4.9t²+50tsin30˚+0

This equation doesn’t factor neatly, so it’d be easier to use the quadratic formula on it.

(-b±√(b2-4ac))/2a
↓
(-50tsin30˚±√((50tsin30˚)2))/-9.8
↓
(-50tsin30˚±(50tsin30˚))/-9.8

So you get 2 answers for t.

0s
(100sin30˚/9.8)s

You have to choose one, and 0s would mean it got launched and landed at the same time, so the second answer must be the right one. Now to use the time the shell stayed in the air to figure out how far it went.

v_x=(50cos30˚)m/s
t=(100sin30˚/9.8)s
↓
s_x=(50cos30˚)m/s•(100sin30˚/9.8)s
↓
s_x=220.924847904193532337684482334933m

So the artillery shell only went about 221 meters, not far enough to hit the camp. There was a cool explosion, though. We’ll just have to come back next time with a more accurate way to aim and a better artillery.

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