The flight path of a projectile is dependent on the angle (theta) that it is fir

Question

The flight path of a projectile is dependent on the angle (theta) that it is fired. This path can be separated into a horizontal and vertical distance as a function of time.

x(t)=t*V_o*cos(theta)

y(t)=t*V_o*sin(theta)-1/2gt^2

whereV_o is the initial velocity, g is the acceleration due to gravity (9.8 m/s2), and t is time in seconds.

The polar coordinates of the projectile at time t are (theta(t), r(t)), where

r(t)=sqrt[x(t)^2 + y(t)^2] and tan(theta)=y(t)/x(t)

Consider the case whereV_o= 162 m/s andtheta= 70

Explanation / Answer

v=162; theta=70*pi/180; % coverting the angle into radian as MATLAB works with radian only. r=zeros(1,31); x=zeros(1,31); y=zeros(1,31); theta1=zeros(1,31); for t=1:31 x(t)=v*cos(theta)*t; y(t)=v*sin(theta)*t-0.5*9.8*t^2; r(t)=sqrt(x(t)^2+y(t)^2); theta1(t)=atan(y(t)/x(t)); end t=1:31; subplot(2,2,1) plot(t,x); title('horizontal distance vs time'); xlabel('time(in sec)'); ylabel('horizontal distance(in m)'); subplot(2,2,2) plot(t,y); title('vertical distance vs time'); xlabel('time(in sec)'); ylabel('vertical distance(in m)'); subplot(2,2,3) plot(x,y); title('vertical distance vs horizontal distance'); xlabel('horizontal distance(in m)'); ylabel('vertical distance(in m)'); subplot(2,2,4) polar(theta1,r); title('polar plot'); xlabel('theta'); ylabel('radius vector');

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