QJEEM24Ph3831: JEE Main 2024 Physics - Class 11 JEEM24Ph3831 Solution: A body projected vertically upwards with a certain speed from…

To solve this problem, we'll use the equations of motion under constant acceleration due to gravity. Let's define:

$h$ : Height of the tower

$u$ : Initial speed of projection (same magnitude in both cases)

$g$ : Acceleration due to gravity (positive downward)

$t_1$ : Time taken to reach the ground when projected upwards

$t_2$ : Time taken to reach the ground when projected downwards

$t$ : Time taken to reach the ground when simply dropped

Case 1: Projectile Thrown Upwards

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:

$h = -u t_1 + \frac{1}{2} g t_1^2$

Case 2: Projectile Thrown Downwards

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:

$h = u t_2 + \frac{1}{2} g t_2^2$

Case 3: Body Dropped

When the body is simply dropped, its initial velocity is $0$ , and it reaches the ground in time $t$ :

$h = \frac{1}{2} g t^2$

Step 1: Equate the Heights

From cases 1 and 2, equate the expressions for $h$ :

$-u t_1 + \frac{1}{2} g t_1^2 = u t_2 + \frac{1}{2} g t_2^2$

Step 2: Solve for $u$

Simplify the equation:

$-u t_1 - u t_2 = \frac{1}{2} g t_2^2 - \frac{1}{2} g t_1^2$

$-u (t_1 + t_2) = \frac{1}{2} g (t_2^2 - t_1^2)$

$-u (t_1 + t_2) = \frac{1}{2} g (t_2 - t_1)(t_2 + t_1)$

Divide both sides by $t_1 + t_2$ :

$-u = \frac{1}{2} g (t_2 - t_1)$

$u = \frac{1}{2} g (t_1 - t_2)$

Step 3: Express $h$ in Terms of $t_1$ and $t_2$

Substitute $u$ back into one of the equations for $h$ :

$h = -\left( \frac{1}{2} g (t_1 - t_2) \right) t_1 + \frac{1}{2} g t_1^2$

Simplify:

$h = \frac{1}{2} g t_1 t_2$

Step 4: Equate the Height for the Dropped Case

From the dropped case:

$h = \frac{1}{2} g t^2$

Set the two expressions for $h$ equal to each other:

$\frac{1}{2} g t^2 = \frac{1}{2} g t_1 t_2$

$t^2 = t_1 t_2$

Step 5: Solve for $t$

$t = \sqrt{t_1 t_2}$

Conclusion:

The time required for the body to reach the ground when dropped is the geometric mean of $t_1$ and $t_2$ :

Answer: Option B

$t = \sqrt{t_1 \, t_2}$