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+ // SAMPLE PROBLEM 3/22
+clc;funcprot(0);
+// Given data
+m=150;// kg
+v_1=4;// m/s
+t_0=0;// s
+t_1=4;// s
+P=600;// N
+t_2=8;// s
+theta=30;// degree
+g=9.81;// The acceleration due to gravity in m/sec^2
+
+// Calculation
+deltat=(m*0)+((m*v_1)-((v_1*2*P)/2)+(m*g*sind(theta)))/((2*P)+(m*g*sind(theta)));// s
+t_a=v_1+deltat;// s
+v_2x=((m*-v_1)+((v_1*2*P)/2)+(v_1*2*P)-(m*g*sind(theta)*t_2))/m;// m/s
+printf("\n(a)The time at which the skip reverses its direction,t_a=%1.2f s \n(b)The velocity of the skip,v_2x=%1.2f m/s",t_a,v_2x);