Answer
c) is correct.
Work Step by Step
According to the principle of conservation of mechanical energy, $$E_f-E_0=0$$ $$\Big(\frac{1}{2}mv_f^2+mgh_f\Big)-\Big(\frac{1}{2}mv_0^2+mgh_0\Big)=0$$ $$\frac{1}{2}m(v_0^2-v_f^2)=mg(h_f-h_0)$$
Let's consider a situation when the ball cannot reach B but only a point C at the same height as A. In this situation, $h_f-h_0=0$, and it follows that $v_0^2-v_f^2=0$, or $v_0=v_f=0$, meaning the ball's initial velocity, or kinetic energy, is 0.
Now, in fact, the ball reaches point B where $h_f-h_0\gt0$, meaning $v_0^2-v_f^2\gt0$, or $v_0\gt v_f=0$. This shows that because of the ball's initial velocity $v_0$, which means it initially has kinetic energy, the ball is able to reach a point higher than its initial position. So c) is correct.
