A body of mass M thrown horizontally with velocity v from the top of the tower of height H touches the ground at a distance of $100 \mathrm{~m}$ from the foot of the tower. A body of mass $2 \mathrm{~M}$ thrown at a velocity $\frac{v}{2}$ from the top of the tower of height $4 \mathrm{H}$ will touch the ground at a distance of _______ m.

<p>To solve this problem, we can use the equations of motion under uniform acceleration, separately considering the horizontal and vertical motions because the two are independent of each other.</p> <p>First, for the body of mass $M$ thrown horizontally with velocity $v$ from a height $H$, let's analyze its motion:</p> <p><strong>Horizontal Motion:</strong><br> <p>The horizontal distance (range) $x$ covered by the object is given by $x = v \cdot t$, where $t$ is the time taken to hit the ground.</p></p> <p><strong>Vertical Motion:</strong><br> <p>The time $t$ it takes for the object to hit the ground can be found using the equation of motion under gravity, $H = \frac{1}{2} g t^2$, where $g$ is the acceleration due to gravity.</p></p> <p><strong>For the first body:</strong><br> <p>Given $x = 100$ m and using the equation for the vertical motion to find $t$, we have:</p> <p>$H = \frac{1}{2} g t^2$</p> <p>Solving for $t$, we get:</p> <p>$t = \sqrt{\frac{2H}{g}}$</p> <p>The horizontal motion gives:</p> <p>$x = v \cdot t \Rightarrow 100 = v \cdot \sqrt{\frac{2H}{g}}$</p></p> <p>Now, considering the second body of mass $2M$ thrown at velocity $\frac{v}{2}$ from height $4H$:</p> <p><strong>For the second body:</strong><br> <p>The time $t'$ it takes for the second body to hit the ground from height $4H$ can be found by:</p> <p>$4H = \frac{1}{2} g t'^2$</p> <p>Solving for $t'$, we get:</p> <p>$t' = \sqrt{\frac{2 \cdot 4H}{g}} = 2 \sqrt{\frac{2H}{g}}$</p> <p>This is twice the time $t$ found for the first body.</p></p> <p>The horizontal distance $x'$ covered by the second body is: <p>$x' = \left(\frac{v}{2}\right) \cdot t'$</p> <p>Now, substituting the value of $t'$ found above, we get:</p> <p>$x' = \left(\frac{v}{2}\right) \cdot 2\sqrt{\frac{2H}{g}} = v \cdot \sqrt{\frac{2H}{g}}$</p> <p>But we previously found that $v \cdot \sqrt{\frac{2H}{g}} = 100$, so:</p> <p>$x' = 100 \text{ m}$</p></p> <p>Therefore, a body of mass $2M$ thrown horizontally with velocity $v/2$ from the top of a tower of height $4H$ will touch the ground at a distance of <strong>100 meters</strong> from the foot of the tower.</p>