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Topic: **Speeding term papers****Question:**
You want to drop a bundle of all your class notes, tests, quizzes, worksheets, and term papers on the 50 yard-line of Lynn C. Adams field during half-time of the first playoff game from a single engine Piper Cub. You fly at a steady height of 522.0 meters and at a speed of 61.0 m/s. How long will it take for the bundle to reach the ground?
And how far in front of the 50 yard-line must the bundle be dropped?

July 22, 2019 / By Dora

(i) Let the time taken to fall 522 m is t sec =>By s = ut + 1/2gt^2 =>522 = 0 + 1/2 x 9.8 x t^2 =>t = √106.53 =>t = 10.32 sec (ii) By R = [Ux] x t =>R = 61 x 10.32 = 629.60 m

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well, you have to check whether the solution is dimensionally correct, meaning that the left side has to have the same units as the right side. You cant have something like X meters = Y seconds. Doesnt make sense. for example, in A, the unit to the left comes out to meters squared, while the units to the right come out to seconds squared.

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The guy above appears correct. The reasoning for this is that x and y motion may be calculated separately in this case... y has acceleration do to gravity, and x only holds the velocity of the aircraft. However, in more complex cases, air resistance must be taken into consideration.

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Neglecting air-resistance: Dy = (½)gt² + (Vyi)t ... initial y_velocity = 0 522 = (½)(9.8)t² t = 10.32 sec ... free-fall time Dx = (Vx) • t Dx = (61) • (10.32) Dx = 630 m ... horizontal distance traveled

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A.) To use Ampere's Law you need to draw an Amperian Loop, which is a rectangle in this case. Here's how to draw it: pretend you are looking along the negative x axis so that J is poking you in the eye. On your paper you would have the positive y axis going horizontally to the right and the positive z axis going up the page. The problem tells you that there's a sheet of current along this y axis and in this new drawing the current is moving out of the page. You can draw the rectangle symmetrically around the origin, so that the origin is at the center. Using intuition and the right-hand-rule for magnetic fields and currents, we know that the horizontal parts of the rectangle will have a B-field parallel to them but that the vertical parts of the rectangle will have the B-field be perpendicular. We now have enough information to use Ampere's Law: Integral of (B dot dl) = (mu naught) * (current enclosed) the above equation is shown graphically here: http://upload.wikimedia.org/math/2/6/e/26e845d2bf8d061986e114bc51ed8254.png The question asks for B as a function of J, so we need to realize another fact: I = J*L, where I is the current enclosed by the rectangle and L is the horizontal length of the rectangle. Evaluating the left hand side of Ampere's Law and substituting J*L for I on the right, we get B*(2L) = (mu naught) * (J*L), so the L's cancel and we get B = 1/2 *(mu naught) * J. There is no x, y, or z dependence in this problem. B.)Substituting J = 2.0 A/m into B = 1/2 *(mu naught) * J, we get B = (mu naught) * (1 A/m). This is the magnitude of B though, not the vector. To get the vector we just need to know the right-hand-rule and realize that at z = .10m the B vector is -(mu naught) * (1 A/m) j-hat. If you plug in the value for mu, you get that the B field is 4pi x 10^-7 (N/(A*m)). Watch the units! Done!

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