A particle is thrown up vertically with a speed `v_(1)` in air . If takes time `t_(1)` in upward journey and `t_(2)(gt t_(1))` in the downward journey and returns to the starting point with a speed `v_(2)`. Then:-

Options

A. `v_(1)=v_(2)`
B. `v_(2) gt v_(1)`
C. `v_(1) gt v_(2)`
D. Data is insufficient

Answer

To solve the problem step by step, we will analyze the motion of the particle during its upward and downward journeys, taking into account the effects of gravity and air resistance.

### Step 1: Analyze the upward journey
When the particle is thrown upwards with an initial speed \( v_1 \), it will experience two forces: the gravitational force acting downwards and the air resistance also acting downwards.

Using the third equation of motion:
\[ v^2 = u^2 + 2as \]
where:
- \( v \) = final velocity at the highest point = 0 (since it momentarily stops)
- \( u \) = initial velocity = \( v_1 \)
- \( a \) = acceleration = \( - (g + a) \) (negative because both gravity and air resistance oppose the motion)
- \( s \) = height reached = \( h \)

Substituting the values, we get:
\[ 0 = v_1^2 - 2(g + a)h \]
Rearranging gives:
\[ v_1^2 = 2(g + a)h \quad \text{(Equation 1)} \]

### Step 2: Analyze the downward journey
During the downward journey, the particle will fall under the influence of gravity, but air resistance will act in the opposite direction (upwards).

Using the same equation of motion:
\[ v^2 = u^2 + 2as \]
where:
- \( v \) = final velocity when it hits the ground = \( v_2 \)
- \( u \) = initial velocity at the highest point = 0
- \( a \) = acceleration = \( g - a \) (net acceleration downwards)
- \( s \) = height fallen = \( h \)

Substituting the values, we get:
\[ v_2^2 = 0 + 2(g - a)h \]
Rearranging gives:
\[ v_2^2 = 2(g - a)h \quad \text{(Equation 2)} \]

### Step 3: Compare the two equations
From Equation 1:
\[ v_1^2 = 2(g + a)h \]
From Equation 2:
\[ v_2^2 = 2(g - a)h \]

To compare \( v_1 \) and \( v_2 \), we can set up the inequality:
\[ v_1^2 = 2(g + a)h \quad \text{and} \quad v_2^2 = 2(g - a)h \]

Since \( g + a > g - a \) (because \( a \) is a positive value representing air resistance), we can conclude:
\[ v_1^2 > v_2^2 \]

Taking the square root of both sides (and noting that both velocities are positive), we have:
\[ v_1 > v_2 \]

### Conclusion
Thus, the relationship between the speeds is:
\[ v_1 > v_2 \]