Need to flip a function and pass it through fminbnd to get the max, function has multiple agruements

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D
D 2011-10-16
I'm trying to use fminbnd to find the max value of a function that has multiple arguements. I know I need to flip the function so the max becomes the minimum, but I can't seem to get the syntax for it, I can't even pass the function through correctly without flipping it.
v0 = 20;
th0 = 45;
h0 = 5;
g = 9.81;
t = linspace(0,4,400);
max = fminbnd(height(t,v0,th0,h0)*(-1),0,4)
%%-----------------------
function [x,y,vx,vy] = trajectory(t,v0,th0,h0,g)
if nargin<=4, g = 9.81; end
if nargin<=3, h0 = 0; end
if nargin==2, th0 = 90; end
th0 = th0*pi/180;
x = v0*cos(th0)*t;
y = h0 + v0*sin(th0)*t - 1/2*g*t.^2;
vx = v0*cos(th0);
vy = v0*sin(th0) - g*t;
%%-------------------------
function y = height(t,v0,th0,h0)
[x,y] = trajectory(t,v0,th0,h0);

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采纳的回答

D
D 2011-10-16
Found the solution with the following code:
tmax = fminbnd(@(t) height(t,v0,th0,h0)*(-1),0,4);
hmax = height(tmax,v0,th0,h0);

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