Question

please solve it using MATLAB

The range of an object shot at an angle theta with respect to the x-axis and an initial velocity v_0 (Figure P3.16) is given by Range = v^2_0/g sin (2 theta) for 0 lessthanorequalto theta lessthanorequalto pi/2 and neglecting air resistance. Use g = 9.81 m/s^2 and an initial velocity v_0 of 100 m/s. Show that the maximum range is obtained at approximately theta = pi/4 by computing the range in increments of pi/100 between 0 lessthanorequalto theta lessthanorequalto pi/2. You won't be able to find the exact angle that results in the maximum range, because your calculations are at evenly spaced angles of pi/100 radian.

Answer 1

clear
clc
g=9.81;
v0=100;
Range=@(theta)v0^2*sin(2*theta)/g;
theta=0:pi/100:2*pi;
All_Range=Range(theta);
[Max,index]=max(All_Range);
fprintf('Theta for which range is maximum:%f\n',theta(index))

Command Window 1lear 2lc Theta for which range is maximum:0.785398 p1/4 g=9.81; v0-100: Range=@ (theta)v0^^*sin (2*theca) /g; theta-0 : p1/ 100: 2*pi ; ans 5 - 0-7854 7Range-Range (theta) -Al1 [Max, index]-max (All_Range); fprintf ('Theta for which range is maximum :%f\n', theta (index)) 9-