{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HWsolution #2

# HWsolution #2 - %Engineering 6 Spring 2004 Problem 2.1...

This preview shows pages 1–2. Sign up to view the full content.

%Engineering 6, Spring 2004, Problem 2.1 Solution %(Golf ball flight) format compact; clc; format short; %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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 4

HWsolution #2 - %Engineering 6 Spring 2004 Problem 2.1...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online