Answer to Question #311041 in MatLAB for OSEGO

Question

A ball is thrown vertically upwards with an initial velocity of 30 m/s.

Using a time step of 0.02 s up to 6.20 s, write a matlab code to give a plot of the vertical distance versus

time for this ball.

Hint ; Motion under gravity is described by the equation : 𝑣𝑦 = 𝑣𝑜𝑦𝑡 +

1

2

𝑔𝑡

2

and gravitational acceleration 𝑔 is here taken as negative.

Then use your code to answer the following questions:

(i) To what maximum height does the ball rise?

(ii) What is the index of time at maximum height?

(iii) How long does it take the ball to ascend to maximum height?

(iv) How long does it take the ball to hit the ground?

(v) What happens to the ball if the sign for gravitational acceleration is taken as positive?

Answer 1

close all,
clear all,
clc,




%{
    A ball is thrown vertically upwards with an initial velocity of 30 m/s.
    Using a time step of 0.02 s up to 6.20 s, write a matlab code to give a plot of the vertical distance versus
    time for this ball.
%}  


u = 30; %Initial Velocity m/s
V0 =30; %m/s, Initial Velocity
t = 0:0.02:6.20;    %Time Step
g = 9.8;    %Gravity g in m/sec^2


%{
S = [];
for r=1:length(t)
    S(r) = V0 - ((1/2)*g*t(r)*t(r));
end
scrsz = get(0,'ScreenSize');
Dim=0;
figure('Position',[scrsz(1)+Dim, scrsz(2)+Dim,scrsz(3)-20,scrsz(4)-100]);
plot(t,S,'r');
grid on,
xlabel('--- Time(t) --->');
ylabel('--- Vertical Distance (meters) --->');
title('Plot: Vertical Distance vs time','FontSize',20);
%}


%(i) To what maximum height does the ball rise? v^2 = u^2 - 2gh
%At maximum height, v = 0, therefore,   h = (u^2/2g)
MaxHeight = (u*u)/(2*g);
fprintf('\n\tMaximum Height = ',MaxHeight);




%(ii) What is the index of time at maximum height?
%h = ut - (1/2)gt^2
A = -(1/2)*g;
B = u;
C = -MaxHeight;
t1 = ((-B)+((B^2-4*A*C))^0.5)/(2*A);
t2 = ((-B)-((B^2-4*A*C))^0.5)/(2*A);
fprintf('\n\tMaximum time to reach the maximum height = %f seconds',t1);
fprintf('\n\tMaximum time to reach the maximum height = %f seconds',t2);


%(iii) How long does it take the ball to ascend to maximum height?
%h = ut - (1/2)gt^2
A = -(1/2)*g;
B = u;
C = -MaxHeight;
t1 = ((-B)+((B^2-4*A*C))^0.5)/(2*A);
t2 = ((-B)-((B^2-4*A*C))^0.5)/(2*A);
fprintf('\n\tHow long does it take the ball to ascend to maximum height? = %f seconds',t1);




%(iv) How long does it take the ball to hit the ground?
fprintf('\n\tMax. time to hit the ground = %f',2*t1);


%(v) What happens to the ball if the sign for gravitational acceleration is taken as positive?
fprintf('\n\nIf g is taken as +ve, that means ball is thrown from top to ground.');