Final answer:
The maximum horizontal distance a man can throw a ball is twice the maximum vertical height if thrown at a 45° angle. Using the projectile motion equations, given the maximum vertical height as 136 m, the maximum horizontal distance is calculated to be 272 m.
Step-by-step explanation:
The maximum vertical height to which a man can throw a ball is given as 136 m. We're trying to find the maximum horizontal distance he can throw the same ball. The maximum distance for a projectile thrown on Earth (ignoring air resistance) is achieved when the ball is thrown at a 45° angle. Given the maximum vertical distance (h), the equation for the maximum height reached by a projectile thrown with initial velocity v is h = (v^2 ÷ 2g), where g is the acceleration due to gravity (9.81 m/s^2).
Since the maximum height is 136 m, we can rearrange the equation to solve for the initial velocity squared which is v^2 = 2gh. Substituting the known values, we get v^2 = 2 × 9.81 m/s^2 × 136 m. Solving for v^2 gives us the initial velocity squared which we can then substitute into the range equation for a projectile R = (v^2 × sin(2°))/g. The sin(90°) is 1, so the equation simplifies to R = v^2/g. By substituting the value of v^2 that we found from the height calculation, it can be shown that the maximum horizontal range is twice the maximum height. Hence, the correct answer is 272 m, which is option D.