Ball Falling On Inclined Ramp

This problem can be solved in the orthodox way, but since this is a Multiple Choice problem, I would prefer (not that much) to use the following method of thinking.

First of all, check the boundary cases.

$\bullet$ When $\theta=0$ ,
Clearly there is going to be collisions at the same point, and hence, the distance between consecutive collision points should be zero .

$\bullet$ When $\theta=\frac{\pi}{2}$ ,
The ball is just going to fall parallel to the wall forever. Hence, there is not going to be any collision at all. Hence, the distance between consecutive collision points should tend to $\infty$ .

$\bullet$ Also, when $n=1$ ,
the problem asks us to find out the distance between the 1st collision point and the 1st collision point, which, clearly is zero .

Analyzing the options, We find that only $4dn(n-1)\tan{\theta}$ satisfies all the above conditions.

NOTE: This might not be a good solution, but I just wanted to show some other unorthodox ways of solving a certain problem. I am currently studying for IIT-JEE, which is a tough exam. And here, it is pretty orthodox to solve a problem this way. Maybe, I will post another solution presenting the correct way of solving this problem. But, anyways, sorry if this isn't good.

Acceleration along the incline is $g\sin\theta$ , and acceleration perpendicular to incline is $g\cos\theta$.

Time taken till the $n^{th}$ collision is equal to $t_n=(2n-1)\sqrt{\frac{2d}{g\cos\theta}}$ So distance covered along incline till the $n^{th}$ collision from initial position is $s=\frac 12 g\sin\theta t_n^2$ distance covered from first collision is $s_1=\frac 12 g\sin\theta (t_n^2-t_1^2)=4dn(n-1)\tan\theta$

vertical distance between first and Nth collision point will be== 4dn(n-1)(sin(theta))^2