Why does a bowling ball roll down faster on a slope than a tennis ball, yet both hits the ground at the same time if they're dropped from the rooftop? If a tennis ball and a bowling ball are dropped of a rooftop, they hit the ground at the same time. But if they are rolled down a slope, the bowling ball rolls faster. Why?
 A: The easy explanation is that the tennis ball is hollow.
When you merely drop the objects, they are subjected to the same acceleration - the aceleration due to gravity - and nothing else. Conservation of energy then says that their gravitational potential energy should be completely transformed into kinetic energy at the ground:
$$mg\Delta h=\frac{1}{2}mv^2\to v=\sqrt{2g\Delta h}$$
Since the initial heights $\Delta h$ are equal, they both have the same velocity as each other (though not constant in time) no matter how far they fall and, thus, hit at the same time.
However, when you roll them down the roof, the initial gravitational potential energy, $mg\Delta h$, is transformed not only into kinetic energy, but also into rotational energy. The rotational energy of something is $\frac{1}{2}I\omega^2$, where $I$ is the moment of inertia (the rotational equivalent of mass) and $\omega$ is the angular velocity ($\omega=v/r$; the velocity of the object divided by its radius).
This is all well and good, so the difference between the bowling ball and the tennis ball is now because the bowling ball is solid and the tennis ball is hollow. When just dropped, there is no difference. However, when rolling, the different distributions of mass affect the moments of inertia differently. A solid sphere has $I=\frac{2}{5}mr^2$, while a hollow sphere (I know the tennis ball is not perfectly hollow, but let's make this approximation, okay?) has $I=\frac{2}{3}mr^2$. What does this mean? Well, let's do the math (math is fun!).
For the bowling ball, we have:
$$mgh=\frac{1}{2}\left(I\omega^2+mv^2\right)=\frac{1}{2}\left(\frac{2}{5}mr^2\cdot\frac{v^2}{r^2}+mv^2\right)\to v=\sqrt{\frac{10}{7}gh}$$
Whereas, for the tennis ball, we have:
$$mgh=\frac{1}{2}\left(I\omega^2+mv^2\right)=\frac{1}{2}\left(\frac{2}{3}mr^2\cdot\frac{v^2}{r^2}+mv^2\right)\to v=\sqrt{\frac{6}{5}gh}$$
Notice that the mass of either ball is mostly irrelevant and that, since $\sqrt{\frac{10}{7}}>\sqrt{\frac{6}{5}}$, the forward velocity, $v$, of the bowling ball is greater than that of the tennis ball; just because one is hollow and one is solid.
It's also worth noting that the radius, as you may have concluded, does not ideally affect the forward velocity. This is something easily shown through the equations above as well as experimentally. Grab some solid spheres of different radii and roll them down an incline (I work in a physics teaching lab, so believe me when I say I've done this many times), you should see they hit the bottom at the same time. Yay! Physics is cool!
A: Ignoring air resistance and other frictional effects other than those causing the objects to roll, the difference is due to the distribution of mass about the axis of rotation and not the actual mass of the two objects.  
The moment of inertia of a body is a measure of the resistance the body to undergo an angular acceleration and the moment of inertia of a solid sphere (bowling ball) is proportionately smaller than that of a hollow sphere (tennis ball) by a factor of about $\frac 3 5$.
This means that the angular acceleration and so also the translational of the bowling ball is greater than that of the tennis ball.  
Put another way, as the bowling ball rolls down a slope proportionately more of the gravitational potential it loses goes into translational kinetic energy and less into the rotational kinetic energy as compared with the energy transfers to a tennis ball.  
The derivation for the acceleration of a solid sphere rolling down a slope is shown to be independent of the mass here and you can adapt the derivation to shown that the acceleration of the bowling ball is greater than that for a tennis ball.  
When he objects fall again their accelerations are independent of mass and since all the loss in gravitational potential energy goes into only translational kinetic energy so the bodies accelerate at the same rate and reach the ground at the same time.
A: I haven't tried this experiment but the first two factors that spring to mind are:


*

*Rolling Friction  The bowling ball is hard and smooth while the tennis ball is fuzzy and softer.  This would lead to a larger coefficient of rolling friction for the tennis ball.

*Distribution of Mass The tennis ball is hollow while the bowling ball is solid.  This means that gram for gram the tennis ball will have a higher moment of inertia which means it takes a greater torque to get it rotate.  See Jim's answer for much more detail.  This is a variation of the classic classroom demo of a solid disc and a ring of equal mass and radius rolling down a ramp - the disk wins.

