# Help With physics problem.

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A projectile was launched perpendicular to a 30deg slope at 20ms^-1. Find the distance between the launching position and landing position.

I tried it many time and keep getting 20âˆš3m, but doesn't match the answers at the back of the book.

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It might be the case that the book is wrong, because I also got the same answer as you do. Are there any other information given? because 20âˆš3m is the answer only if we assume that the landing place is of the same height as the launching place.

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Actually that was my assumption too. However it doesn't make any sense if it was on ground level and has slope of 30. They haven't stated anything about the hight of which it was thown from.

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that's really weird then. I can't think of any other reasons why the answer is not 20âˆš3m using the given information. I think we should blame for the vague wording of the question then

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May be a dumb question, but how did you get 20*sqrt(3)? Could anyone please explain?

From the diagram, it's clear that the vertical velocity is 20sin(60), and the horizontal velocity is 20cos(60)

Now assume that the projectile lands at the same height as where it was launched, then the vertical velocity of the projectile when it lands is -20sin(60) (the minus sign represents the opposite direction of the velocity compared to when it was launched). Also let g = -10 m/s^2 stand for acceleration due to gravity.

Now we substitute all the information into the formula v = u + at, we get -20sin(60) = 20sin(60) -10t, which yields the total time taken to be t = 2âˆš3

Now using the horizontal velocity and the formula s = vt, we get the distance to be s = 20cos(60) * 2âˆš3 = 20âˆš3 m. And that is the answer

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