HW2 #1 - %y = vy * (x/vx) - (1/2) * g * (x/vx)^2 %y = x *...

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Sheet1 Page 1 %Engineering 6, Spring 2004, Problem 2.1 %Ryan Uy, 2509 %Section B05, Tues. 3-4 %Given variables: angle = 14 degrees, distance = 325 yards, roll = 15 yards %gravity = 32.2 feet per second squared %convert distance in x direction from yards to feet and compensate for roll x = 325 * 3 - 15 %convert angle from degrees to radians a = 14 * (pi / 180) g = 32.2 %distance in y direction given by following equation: %y = initial position (0) + initial velocity (vy) * time (t) - (1/2) * acceleration (g) * time (t) squared %y = vy * t - (1/2) * g * t^2 %distance in x direction given by following equation: %x = initial velocity (vx) * time (t) %x = vx * t %solve for time (x-distance equation): %t = x / vx %substitute time into y-distance equation:
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Unformatted text preview: %y = vy * (x/vx) - (1/2) * g * (x/vx)^2 %y = x * (vy/vx) - (1/2) * g * (x/vx)^2 %tan (a) = vy / vx %y = x * tan (a) - (1/2) * g * (x/vx)^2 %solve for vx when y=0: vx = x / sqrt((2/g) * x * tan(a)) %find value for t: t = x / vx %solve for vy: vy = vx * tan(a) %find maximum height (y value) by setting t=t/2 (since trajectory is symmetrical) y = vy * (t/2) - (1/2) * g * (t/2)^2 %solve for initial velocity (vi): vi = vx / cos(a) %convert intial velocity from feet per second to miles per hour: vi = vi *(1/5280) * 60 * 60 %display quantities fprintf('Initial Velocity = %.1f miles per hour\n', vi) Sheet1 Page 2 fprintf('Maximum Height = %.1f feet\n', y) fprintf('Time of Flight = %.1f seconds\n', t)...
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HW2 #1 - %y = vy * (x/vx) - (1/2) * g * (x/vx)^2 %y = x *...

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