A body projected vertically upwards with a certain speed from the top of a tower reaches the ground in $t_1$. If it is projected vertically downwards from the same point with the same speed, it reaches the ground in $t_2$. Time required to reach the ground, if it is dropped from the top of the tower, is :

<p>To solve this problem, we'll use the equations of motion under constant acceleration due to gravity. Let's define:</p> <p><p>$ h $: Height of the tower</p></p> <p><p>$ u $: Initial speed of projection (same magnitude in both cases)</p></p> <p><p>$ g $: Acceleration due to gravity (positive downward)</p></p> <p><p>$ t_1 $: Time taken to reach the ground when projected upwards</p></p> <p><p>$ t_2 $: Time taken to reach the ground when projected downwards</p></p> <p><p>$ t $: Time taken to reach the ground when simply dropped</p></p> <p><strong>Case 1: Projectile Thrown Upwards</strong></p> <p>When the body is projected upwards from the top of the tower, its initial velocity is $ -u $ (since upward direction is negative), and it reaches the ground in time $ t_1 $. The equation of motion is:</p> <p>$ h = -u t_1 + \frac{1}{2} g t_1^2 $</p> <p><strong>Case 2: Projectile Thrown Downwards</strong></p> <p>When the body is projected downwards from the top of the tower, its initial velocity is $ u $, and it reaches the ground in time $ t_2 $. The equation is:</p> <p>$ h = u t_2 + \frac{1}{2} g t_2^2 $</p> <p><strong>Case 3: Body Dropped</strong></p> <p>When the body is simply dropped, its initial velocity is $ 0 $, and it reaches the ground in time $ t $:</p> <p>$ h = \frac{1}{2} g t^2 $</p> <p><strong>Step 1: Equate the Heights</strong></p> <p>From cases 1 and 2, equate the expressions for $ h $:</p> <p>$ -u t_1 + \frac{1}{2} g t_1^2 = u t_2 + \frac{1}{2} g t_2^2 $</p> <p><strong>Step 2: Solve for $ u $</strong></p> <p>Simplify the equation:</p> <p>$ -u t_1 - u t_2 = \frac{1}{2} g t_2^2 - \frac{1}{2} g t_1^2 $</p> <p>$ -u (t_1 + t_2) = \frac{1}{2} g (t_2^2 - t_1^2) $</p> <p>$ -u (t_1 + t_2) = \frac{1}{2} g (t_2 - t_1)(t_2 + t_1) $</p> <p>Divide both sides by $ t_1 + t_2 $:</p> <p>$ -u = \frac{1}{2} g (t_2 - t_1) $</p> <p>$ u = \frac{1}{2} g (t_1 - t_2) $</p> <p><strong>Step 3: Express $ h $ in Terms of $ t_1 $ and $ t_2 $</strong></p> <p>Substitute $ u $ back into one of the equations for $ h $:</p> <p>$ h = -\left( \frac{1}{2} g (t_1 - t_2) \right) t_1 + \frac{1}{2} g t_1^2 $</p> <p>Simplify:</p> <p>$ h = \frac{1}{2} g t_1 t_2 $</p> <p><strong>Step 4: Equate the Height for the Dropped Case</strong></p> <p>From the dropped case:</p> <p>$ h = \frac{1}{2} g t^2 $</p> <p>Set the two expressions for $ h $ equal to each other:</p> <p>$ \frac{1}{2} g t^2 = \frac{1}{2} g t_1 t_2 $</p> <p>$ t^2 = t_1 t_2 $</p> <p><strong>Step 5: Solve for $ t $</strong></p> <p>$ t = \sqrt{t_1 t_2} $</p> <p><strong>Conclusion:</strong></p> <p>The time required for the body to reach the ground when dropped is the geometric mean of $ t_1 $ and $ t_2 $:</p> <p><strong>Answer:</strong> Option B</p> <p>$ t = \sqrt{t_1 \, t_2} $</p>