Physics

Vertical motion.  Distance given by:  s_vert = 11.5*sin(35°)*t - (1/2)*9.81*t²

 This is zero at times t = 0 and t = 2*11.5*sin(35°)/9.81 seconds.  The latter time is the time at which the rock comes back to the ground.

Horizontal motion.   Distance given by:  s_horiz = 11.5*cos(35°)*t

So, total horizontal distance travelled = 11.5*cos(35°)*2*11.5*sin(35°)/9.81 metres

I'll leave you to crunch the numbers.

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It dependes on how hevvy the rock is.

The weight of the rock dosen't change anything, as the initial velocity is already known and acceleration due to gravity is the same for all objects on Earth no matter what the mass is. The question ignores air resistance.

If you just want the horizontal distance, here’s a quick way.

Distance equals the square of the velocity divided by the acceleration of gravity times the sine of twice the angle. 

\(R = \dfrac {V^2 }{g} \times sin (2\theta)\)

This is an easy way to check the distance result for questions requiring more complex solutions. 
 

\(\dfrac {11.5^2 }{9.81} \times sin (2*35)=12.67 \text { meters}\)

This matches Sir Alan’s solution for distance.