Please help to spot the error here

This is a solved example in the book 'Applied numerical methods with MATLAB by Chapra'

A bungee jumper jumps off a cliff 200 m high with an upward velocity of 20 m/s. Detemine when he hts the ground and plot dist vs time and velocity vs time.

The problem is solved in the book as below : Assuming DOWNWARD direction is positive (i.e. distance (y), velocity (v) and force are positive downards, and x = 0 at ground levele. The boundary conditions are : x(0) = -200 m/s, v(0) = -20 m/s

The diff eqn being solved is : md2y/dt2 = Fdown - Fup = mg - cd / m * v^2

The following functions are written

1. Function for derivative --- Here

function dydt=freefall(t,y,cd,m)
% y(1) = x and y(2) = v
grav=9.81;
dydt=[y(2);grav-cd/m*y(2)*abs(y(2))];
end

2. Function to detect event of hitting the ground

    % Event function
 function [detect,stopint,direction]=endevent(t,y,varargin)
% Locate the time when height passes through zero
% and stop integration.
detect=y(1); % Detect height = 0
stopint=1; % Stop the integration
direction=0; % Direction does not matter
end

3. Script file to run the problem

    % Script file
  opts=odeset('events',@endevent);
y0=[-200 -20];
[t,y,te,ye]=ode45(@freefall,[0 inf],y0,opts,0.25,68.1);
te,ye
plot(t,y(:,1),'-',t,y(:,2),'--','LineWidth',2)
legend('Height (m)','Velocity (m/s)')
xlabel('time (s)');
ylabel('x (m) and v (m/s)')
end

The answer for this is : jumper hits the ground after 9.5475 s at 46.2454 m/s (downward)

==================================================================================== I am trying to solve THE SAME problem using the following convention :

Assuming UPWARD direction is positive (i.e. distance (y), velocity (v) and force are positive downards, and x = 0 at ground levele. the boundary conditions are : x(0) = 200 m, v(0) = 20 m/s

The diff eqn being solved is : md2y/dt2 = -Fdown + F up = -mg + cd / m * v^2

The following functions are written

1. Function for derivative

    % code
 function dydt=freefall(t,y,cd,m)
% y(1) = x and y(2) = v
grav=9.81;
dydt=[y(2);-grav + cd/m*y(2)*abs(y(2))];
end

2. Function to detect event of hitting the ground

% code
function [detect,stopint,direction]=endevent(t,y,varargin)
% Locate the time when height passes through zero
% and stop integration.
detect=y(1); % Detect height = 0
stopint=1; % Stop the integration
direction=0; % Direction does not matter
end

3. Script file to run the problem

% code
opts=odeset('events',@endevent);
y0=[200 20];
[t,y,te,ye]=ode45(@freefall,[0 inf],y0,opts,0.25,68.1);
te,ye
plot(t,y(:,1),'-',t,y(:,2),'--','LineWidth',2)
legend('Height (m)','Velocity (m/s)')
xlabel('time (s)');
ylabel('x (m) and v (m/s)')
end

The answer for this is : jumper hits the ground after 8.0142 s at -104.8502 m/s (downward)

I am unable to spot the error. Will someone PLEASE explain where the mistake is ?

TIA