Question 15: Motion in 1D

Question

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 :

Accepted Answer

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} $

Answer

$$\sqrt{\mathrm{t}_1+\mathrm{t}_2}$$

Answer

$$\sqrt{\mathrm{t}_1 \mathrm{t}_2}$$

Answer

$$\sqrt{\frac{\mathrm{t}_1}{\mathrm{t}_2}}$$

Answer

$$\sqrt{\mathrm{t}_1-\mathrm{t}_2}$$

Motion in 1D

Choose the correct answer: