The maximum vertical height to which a man can throw a ball is 136 m. The maximum horizontal distance upto which he can throw the same ball is

Let the maximum speed with which the man can release the ball be $$u$$.

Case 1: Ball thrown vertically upward.
Maximum height reached is given by the kinematic formula

$$h_{\max}= \frac{u^{2}}{2g}$$

The question states that $$h_{\max}=136\; \text{m}$$. Hence

$$\frac{u^{2}}{2g}=136$$

$$\Rightarrow u^{2}=2g \times 136 = 272\,g \qquad -(1)$$

Case 2: Ball thrown for maximum horizontal range.
For a given speed $$u$$, the horizontal range $$R$$ of a projectile is

$$R=\frac{u^{2}\sin 2\theta}{g}$$

The range becomes maximum when $$\theta = 45^{\circ}$$ because $$\sin 90^{\circ}=1$$. Therefore, the maximum range is

$$R_{\max}= \frac{u^{2}}{g} \qquad -(2)$$

Substitute $$u^{2}$$ from equation $$(1)$$ into $$(2)$$:

$$R_{\max}= \frac{272\,g}{g}=272\;\text{m}$$

Thus the farthest horizontal distance the man can throw the ball is $$272\;\text{m}$$.

Hence, Option C is correct.