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

We know that the maximum vertical height a man can throw a ball is 136 m, and our goal is to determine the maximum horizontal distance.

First, by finding the initial speed from the maximum height, we note that the height is reached when the ball is thrown vertically upward ($$\theta = 90°$$). Using the standard relation for maximum height, we have:

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

Solving for $$u^2$$ gives $$u^2 = 2g \times h_{max} = 2 \times 10 \times 136 = 2720\;\text{m}^2/\text{s}^2$$, so the square of the initial speed is 2720.

Next, the maximum horizontal range on flat ground occurs at a launch angle of $$\theta = 45°$$, yielding:

$$ R_{max} = \frac{u^2 \sin(2 \times 45°)}{g} = \frac{u^2 \sin 90°}{g} = \frac{u^2}{g} $$

Substituting $$u^2 = 2720$$ and $$g = 10\;\text{m}/\text{s}^2$$ gives:

$$ R_{max} = \frac{2720}{10} = 272\;\text{m} $$

We also note the useful general result that $$R_{max} = 2h_{max}$$. Therefore, the maximum horizontal distance is 272 m, which corresponds to Option 3: 272 m.