Question

2. The trajectory motion of a projectile is given as a function of time. x(t) = (V, cost 1 y(t) = (v, sin(O)t - 5 gta where x is the horizontal distance or down range of the projectile in feet y is the vertical motion or height of the projectile in feet ve is the initial velocity = 200 ft/s O is the angle with the horizontal when the projectile was launched g is the acceleration due to gravity [ft/s?] (Use the correct value for g) a. Derive an expression to find the total time of flight tq, t2, and tz corresponding to angles of eat 30°, 37.5°, and 45° respectively. Use MATLAB to solve the equation and present the results for the total time of flight for ty, ty, and tą on the command window with 3 decimal places of accuracy. Hint: Think about the value of y(t) when the projectile reaches the total time of flight! b. On a new figure window, plot the three curves corresponding to xi(t) versus t, xz(t) versus t, and X3(t) versus t. These three curves should be in the same figure window. Use 500 evenly spaced points to generate each of the curves. Where X1= horizontal motion of the projectile when 0 = 30° during the time interval 0 <tsti X = horizontal motion of the projectile when 0 = 37.5º during the time interval 0 <t<t Xz= horizontal motion of the projectile when 6 = 45° during the time interval 0 <t<ta On a new figure window, plot three curves corresponding to yı(t) versus t, yz(t) versus t, and yz(t) versus t. These three curves should be in the same figure window. Use 500 evenly spaced points to generate each of the curves. Where Vi= vertical motion of the projectile when 0 = 30° during the time interval 0 <t<ti V = vertical motion of the projectile when = 37.5º during the time interval 0 <t<t V = vertical motion of the projectile when 6 = 45° during the time interval 0 <t<tz d. On a new figure window, plot three curves corresponding to y.(t) versus xi(t), yz(t) versus xz(t), and yzt) versus xzt) to show the projectile motions in the x - y plane. These three curves should be in the same figure window.

I have no idea what to do. Can you please include the code with it? Thanks!

Answer 1

clc
clear all

g = 32.17; % ft/s^2
v0 = 200; % ft/s
thetad = [30 37.5 45 ]; % theta in degree
theta = deg2rad(thetad); % theta in radian
color = ['r','k','b'];

%% --(a)--
% tf = time of flight
% at t =tf, projectile will fall back to ground
% Hence, to find tf, we have to evalueate y = 0 such taht t>0 (root other
% than t = 0
syms t
for i =1:length(theta)
y = v0*sin(theta(i))*t-0.5*g*t^2;
tf(i) = solve([y==0,t>0],t);
end
tf = double(tf); % converting sym to double
clc
for i = 1:length(theta)
fprintf('time of flight tf%i = %0.3f second\n',i,tf(i));
end
%% --(b)--
x = [];
figure
hold on
for i =1:length(thetad)
t = linspace(0,tf(i),500);
x = [x; v0*cos(theta(i))*t];
plot(t,x(i,:),color(i));
end
grid on
xlabel('time(s)');
ylabel('horizantal distance (ft)');
legend('x_1(t)','x_2(t)','x_3(t)','Location','north');
%% --(c)--
y = [];
figure
hold on
for i =1:length(thetad)
t = linspace(0,tf(i),500);
y = [y; v0*sin(theta(i))*t-0.5*g*t.^2];
plot(t,y(i,:),color(i));
end
grid on
xlabel('time(s)');
ylabel('vertical distance (ft)');
legend('y_1(t)','y_2(t)','y_3(t)');
ylim([0 350]);
%% --(d)--
figure
hold on
for i =1:length(theta)
plot(x(i,:),y(i,:),color(i));
end
grid on
xlabel('x(t) ft');
ylabel('y(t) ft');
legend('\theta_1=30 degree','\theta_2=37.5 degree','\theta_3=45 degree');
ylim([0 350]);

Command Window time of flight tfl = 6.217 second time of flight tf2 = 7.569 second time of flight tf3 = 8.792 second

350 . 0,=30 degree 8=37.5 degree 6,=45 degree 300 ...... 250 ....... y(t) ft 50 ...../. 200 400 600 800 1000 1200 1400 x(t) ft

1400 *g(t) 1200 ...... xz(t) 1000 ....... horizantal distance (ft) 200 ....... 2 w time(s)

350 . 300 ...... 250 ....... vertical distance (ft) 100 ... 50 .....// w time(s)