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# Projectile on an incline

Challenging problem of a projectile on an inclined plane. Created by Sal Khan.

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• I have a big PROBLEM from .Lets say the angel of hill was something much more complicated then 30 degrees,like 73.486 degrees, then we would'nt know the ratio of Sy to Sx...there must be another way...is there ???? •   If you don't have a calculator (or sin/cos/tan tables) handy, 73 degrees will be more problematic than 30. If you have a protractor you can draw the triangle with angle 73.something degrees, and measure out SOH CAH TOA using a ruler. You will be limited by the accuracy of your drawing, though!

Some of the most common and useful functions are difficult to compute by hand. Before the calculator, mathematicians (and students) used reference tables of precomputed values.
Logarithms were built on purpose to work that way. Some poor guy spends his whole life building tables, and from then on, the rest of us can just look the numbers up.
If what you meant is "what would I do all by myself?", I think that is a very good question!

For sample exercises, you often use values like 30 degrees because they're easy to work with. When we get the concept, the steps will be the same in a real situation. Some of the intermediate steps may be more tedious, that's all.
• Instead of having a tilted plane that you are shooting off of, couldn't you tilt everything by 30 degrees, making the tilted plane flat, and gravity pulling at a 60 degree angle instead of a 90? All you would have to do is use trig to find that instead of gravity having a force of -9.8, it had a force of -4.9 square root of 3.
Would this work? • If you factor in air resistance, how would that change the problem? And how would you solve it? • Can we assume our frame of reference with respect to the inclined plane???
I mean.. our inclined plane will be the x axis.. in this case.. the condition for the tootal distance travelled will be y co-ordinate will be 0...????
Then we can simply use projectile motion formulas by using theta as angle with ground -angle of inclined plane with ground??? • Why don't we take the incline as the horizontal axis? That way the horizontal displacement alone should be our answer, shouldn't it? • I solve for time with tan (30) = (7.07t-4.9t²)/(7.07t) => t=0.83

But when I want to figure out the horizontal component Sx then:
7.07*t = Sx
But: 7.07*0.83 = 5.86
and Sal has 4.31
I don't understand why • Can anyone explain how we could do it if we take the same hill but we fire the porjectile from its top? With the same angle as before in the example above. • At around why does the first S sub y end up becoming 1? Thank you • The earlier equation was:
S sub y=sqrt(3)S sub y - 4.9 (3*S sub y/50)
Now subtracting S sub y on both sides:
0 = sqrt( 3)S sub y - S sub y - 4.9(3*S sub y/50)

In the expresion - (sqrt(3)S sub y -S sub y), we can take S sub y common...
so it becomes,

0 = S sub y
(sqrt(3) -1) - 4.9*(3*S sub y / 50)

Got it!!

Thank You  