creating a function to find horizontal distance given initial velocity and initial angle.
이전 댓글 표시
Im trying to create a function that will allow for an input variable of the initial velocity, v and the initial angle, a; that comes up with the output of the horizontal distance once the object gets back to the starting y position.
so far I have
v = input('enter cannon ball exit velocity(m/s):'); %input of the exit velocity of the cannon in m/s a = input('enter cannon ball exit angle from horizon(degrees):'); %input of the exit angle from the horizon of the cannon ball in degrees
function [x]=cannon(v,a) g = -9.8; %gravitaitional constant in m/s^2 t = (2*(v*sin(a)))/g; %equation for the time until cannon hits the ground at the inpact location x = (v*cos(a))*t %equation for the horizontal distance where the cannon hits the ground at the impact location end
댓글 수: 2
Dimitris Kalogiros
2018년 8월 6일
편집: Dimitris Kalogiros
2018년 8월 6일
1) what is the quetion ?
Adam Danz
2018년 8월 7일
Is this a coding question or a conceptual question?
답변 (1개)
Adam Danz
2018년 8월 7일
It's not clear whether your question is related to code or concept. Concerning the code, you pretty much already have it written. I filled in some lines and reformatted it so that your function is called after the inputs are collected and it produces an output printed in the command window.
v = input('enter cannon ball exit velocity(m/s):'); %input of the exit velocity of the cannon in m/s
a = input('enter cannon ball exit angle from horizon(degrees):'); %input of the exit angle from the horizon of the cannon ball in degrees
x = cannon(v,a);
fprintf('The horizontal displacement is %.2f.\n', x)
function x = cannon(v,a)
g = -9.8; %gravitaitional constant in m/s^2
t = (2*(v*sin(a)))/g; %equation for the time until cannon hits the ground at the inpact location
x = (v*cos(a))*t %equation for the horizontal distance where the cannon hits the ground at the impact location end
end
I haven't looked at the code conceptually.
카테고리
도움말 센터 및 File Exchange에서 Geology에 대해 자세히 알아보기
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!