The line $x-2y-1=0$ insects the parabola $y^2=4x$ at two different points $A, B$. Let $C$ be a point on the parabola such that $\angle ACB=\frac{\pi}{2}$. Find the coordinate of point $C$.
1. **Find the points of intersection of the line and the parabola:**
   The line equation is \( x - 2y - 1 = 0 \). We can express \( x \) in terms of \( y \):
   \[
   x = 2y + 1
   \]
   Substitute this into the parabola equation \( y^2 = 4x \):
   \[
   y^2 = 4(2y + 1)
   \]
   Simplify and solve for \( y \):
   \[
   y^2 = 8y + 4
   \]
   \[
   y^2 - 8y - 4 = 0
   \]
   Solve this quadratic equation using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
   \[
   y = \frac{8 \pm \sqrt{64 + 16}}{2} = \frac{8 \pm \sqrt{80}}{2} = \frac{8 \pm 4\sqrt{5}}{2} = 4 \pm 2\sqrt{5}
   \]
   Thus, the \( y \)-coordinates of the intersection points are \( y_1 = 4 + 2\sqrt{5} \) and \( y_2 = 4 - 2\sqrt{5} \).

2. **Find the corresponding \( x \)-coordinates:**
   Using \( x = 2y + 1 \):
   \[
   x_1 = 2(4 + 2\sqrt{5}) + 1 = 9 + 4\sqrt{5}
   \]
   \[
   x_2 = 2(4 - 2\sqrt{5}) + 1 = 9 - 4\sqrt{5}
   \]
   Therefore, the points of intersection are \( A(9 + 4\sqrt{5}, 4 + 2\sqrt{5}) \) and \( B(9 - 4\sqrt{5}, 4 - 2\sqrt{5}) \).

3. **Find the coordinates of point \( C \) on the parabola such that \( \angle ACB = \frac{\pi}{2} \):**
   Let \( C(t^2, 2t) \) be a point on the parabola \( y^2 = 4x \). For \( \angle ACB = \frac{\pi}{2} \), the slopes of \( AC \) and \( BC \) must satisfy:
   \[
   m_{AC} \cdot m_{BC} = -1
   \]
   Calculate the slopes:
   \[
   m_{AC} = \frac{2t - (4 + 2\sqrt{5})}{t^2 - (9 + 4\sqrt{5})} = \frac{2t - 4 - 2\sqrt{5}}{t^2 - 9 - 4\sqrt{5}}
   \]
   \[
   m_{BC} = \frac{2t - (4 - 2\sqrt{5})}{t^2 - (9 - 4\sqrt{5})} = \frac{2t - 4 + 2\sqrt{5}}{t^2 - 9 + 4\sqrt{5}}
   \]
   Set up the product of the slopes:
   \[
   \left( \frac{2t - 4 - 2\sqrt{5}}{t^2 - 9 - 4\sqrt{5}} \right) \left( \frac{2t - 4 + 2\sqrt{5}}{t^2 - 9 + 4\sqrt{5}} \right) = -1
   \]
   Simplify the product:
   \[
   \frac{(2t - 4)^2 - (2\sqrt{5})^2}{(t^2 - 9)^2 - (4\sqrt{5})^2} = -1
   \]
   \[
   \frac{(2t - 4)^2 - 20}{(t^2 - 9)^2 - 80} = -1
   \]
   Simplify further:
   \[
   (2t - 4)^2 - 20 = -(t^2 - 9)^2 + 80
   \]
   \[
   4t^2 - 16t + 16 - 20 = -t^4 + 18t^2 - 81 + 80
   \]
   \[
   4t^2 - 16t - 4 = -t^4 + 18t^2 - 1
   \]
   \[
   t^4 - 14t^2 - 16t - 3 = 0
   \]
   Solve this polynomial equation:
   \[
   (t + 1)(t + 3)(t^2 - 4t - 1) = 0
   \]
   The roots are \( t = -1, -3 \) and the roots of \( t^2 - 4t - 1 = 0 \).

4. **Find the coordinates of \( C \):**
   For \( t = -1 \):
   \[
   C(1, -2)
   \]
   For \( t = -3 \):
   \[
   C(9, -6)
   \]

The final answer is \( \boxed{(1, -2) \text{ or } (9, -6)} \).