ps2solnsW06 - %Engineering 6 Winter 2006 Problem 2.1...

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%Engineering 6, Winter 2006, Problem 2.1 Solution %(Golf ball flight) %Clear command window and variables clc; clear; %The equation for the distance the ball travels is: % Total distance = distance in air + roll %In lecture and/or the notes (p. 6), it was derived that % the time of flight is 2vsin(theta)/g and the air distance % is the time of flight times vcos(theta). We can therefore % solve for the velocity, since we know everything else: % velocity^2 = (g/(2sin(theta)cos(theta)))(total_dist - roll_dist). %We will do the calculation in units of feet and seconds, and % then convert it to miles per hour at the end %Define variables theta = 14*pi/180; %14 degrees in radians total_dist = 325*3; %Total distance in feet roll_dist = 15*3; %Roll distance in feet g = 32.2; %Acceleration due to gravity in ft/sec^2 %Calculate velocity in feet per second (but first subtract %roll distance to get distance flown in the air) air_dist = total_dist - roll_dist; v = sqrt(g*air_dist/(2*sin(theta)*cos(theta))); %Convert velocity to miles per hour v_mph = (3600/5280)*v; %Calculate time of flight t_f = 2*v*sin(theta)/g; %Calculate max height reached (in feet) %Since the ball's trajectory in the air is symmetrical (because % we're ignoring the ball's spin, etc.), the time to reach
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This note was uploaded on 11/02/2009 for the course EEC eng 6 taught by Professor Lagerstrom during the Winter '06 term at UC Davis.

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ps2solnsW06 - %Engineering 6 Winter 2006 Problem 2.1...

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