How many times will a ball spin if I drop it of a height $h$? I'm currently doing an extended project qualification and my topic is to form an equation which maximizes the probability of the ball landing on a point marked with a red pen on the ball.
Assuming that the ball is exactly spherical, with a diameter of 'd'. The ball is rolled of a curved downwards surface to make friction negligible . The ball moves down a vertical distance of height H_t  and is in contact with the  surface for a distance of twice the circumference of the ball. It starts rotating when the red dot touches the surface and rolls off when the red dot touches the surface. So as sum of energy is conserved KE = GPE V_final when the ball leaves the surface is V_final = sqrt(2gH_t). The ball then undergo projectile motion with horizontal speed of sqrt(2gH_t) and moves a horizontal distance of R. And it moves down a vertical distance of H_f as it reaches the ground. Using the idea of Torque = rotational inertia x angular acceleration is it possible to determine how many times the ball will spin before it touches the ground. or for a time of T seconds which can be calculated by the verticle distance. 
I need some recommendations of physics concepts I should look at for example, newton's second law of rotation, and etc to be able to determine this. Also, i want to apply air resistance to the system.
Given area of the sphere incontact with the air is,
pi x d^2 ,
Drag coefficient is;
C_d
Density of air is ;
Rho
My plan is once i get torque = rotational inertia x angular acceleration, I can integrate both sides twice to get angular displacement setting boundary conditions
 A: I recommend that you neglect air resistance at this stage. It reduces linear and rotational speed, but the effect should be small for a dense, smooth ball.
First you need to work out the rotational speed $\omega$ of the ball, which will be constant as it falls. (There is no torque acting on the ball while it is in flight, if drag is neglected.) Assuming the ball is rolling and not slipping at the point of launch, $\omega$ is related to the linear speed $v$ of the CM at the point of launch by $v=\omega d$. You also need the time of flight $T$. 
Both $v$ and $T$ can be worked out from the height $H$ and range $R$ of the projectile. 
If you do not know the range $R$, you can calculate $v$ and $\omega$ from the potential energy lost as the ball rolls down the slope. Assuming the ball does not slip as it rolls down the slope, you can use conservation of energy - as you have attempted to do, but you have neglected rotational KE. So you should have
$mgH_1=\frac12mv^2+\frac12J\omega^2$
where $J=\frac25md^2$ is the moment of inertia about the centre. However, this calculation is not as reliable as using the range $R$, because it assumes there is no slipping at any point on the slope, whereas using $R$ only assumes no slipping at the launch point.
Finally, to land on the red spot the ball must complete a whole number $n$ of revolutions in time of flight $T$ : ie $n=(\omega/2\pi)T$. 
