A person can throw a ball upto a maximum range of 100 m. How high above the ground he can throw the same ball?

<p>To determine how high a person can throw a ball given the maximum range, we need to use the principles of projectile motion in physics. The maximum range of a projectile is given by the formula:</p> <p>$R = \frac{{v_0^2 \sin(2\theta)}}{g}$</p> <p>where:</p> <ul> <li>$R$ is the range</li> <li>$v_0$ is the initial velocity</li> <li>$\theta$ is the angle of projection</li> <li>$g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$)</li> </ul> <p>The maximum range is achieved when $\theta = 45^\circ$, thus $\sin(2\theta) = \sin(90^\circ) = 1$.</p> <p>Given that the maximum range $R = 100 \, \text{m}$, we can rewrite the range formula as:</p> <p>$100 = \frac{{v_0^2 \cdot 1}}{9.8}$</p> <p>Solving for $v_0^2$:</p> <p>$v_0^2 = 100 \times 9.8 = 980$</p> <p>Next, the maximum height $H$ reached by the ball can be calculated using the vertical component of the velocity. The formula for the maximum height is:</p> <p>$H = \frac{{v_0^2 \sin^2(\theta)}}{2g}$</p> <p>Here, for a vertical throw, $\theta = 90^\circ$, and thus $\sin(90^\circ) = 1$.</p> <p>Substituting the values, we get:</p> <p>$H = \frac{{v_0^2 \cdot 1}}{2 \cdot 9.8}$</p> <p>Using $v_0^2 = 980$, we have:</p> <p>$H = \frac{980}{2 \times 9.8} = \frac{980}{19.6} = 50 \, \text{m}$</p> <p>Therefore, the correct answer is:</p> <p>Option B: 50 m</p>