Two objects A and B are thrown upward simultaeously with the same speed. The mass oif A is greater than the mass of B. Suose the ir exerts a constant and equal force of resitance on the two bodies.

Options

A. The two bodies will rech the same height
B. A will go higher than B.
C. B will go higher than A
D. Any of the above three may happen depending on the speed with which the objects are thrown

Answer

To solve the problem of which object, A or B, will go higher when thrown upward with the same initial speed, we can analyze the forces acting on both objects and use Newton's laws of motion. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the Forces Acting on Each Object
Both objects A and B experience the following forces when thrown upwards:
- The gravitational force acting downwards:
- For object A: \( F_g = m_A \cdot g \)
- For object B: \( F_g = m_B \cdot g \)
- The air resistance (drag force) acting downwards, which is equal for both objects:
- \( F_r \)

### Step 2: Write the Net Force Equations
Using Newton's second law, we can write the net force equations for both objects:
- For object A:
\[ F_{net, A} = F_r + m_A \cdot g = m_A \cdot a_A \]
- For object B:
\[ F_{net, B} = F_r + m_B \cdot g = m_B \cdot a_B \]

### Step 3: Rearranging the Equations
Rearranging the equations gives us the expressions for acceleration:
- For object A:
\[ a_A = \frac{F_r + m_A \cdot g}{m_A} \]
- For object B:
\[ a_B = \frac{F_r + m_B \cdot g}{m_B} \]

### Step 4: Compare the Accelerations
Since \( m_A > m_B \) and \( F_r \) is constant for both, we can see that:
- The acceleration of A will be less than the acceleration of B because the same force (air resistance) is acting on a larger mass (A). Therefore, we can conclude:
\[ a_A < a_B \]

### Step 5: Analyze the Motion
Both objects are thrown with the same initial velocity \( v \) upwards. The object with the lesser downward acceleration (which is object A) will decelerate less and thus will rise higher before coming to a stop.

### Step 6: Use the Equation of Motion
We can also use the equation of motion to find the maximum height:
\[ v^2 = u^2 + 2a s \]
Where:
- \( v \) is the final velocity (0 at the maximum height),
- \( u \) is the initial velocity,
- \( a \) is the acceleration (which is negative in this case),
- \( s \) is the height.

Rearranging gives:
\[ s = \frac{v^2}{2(-a)} \]
Since \( a_A < a_B \), it follows that:
\[ s_A > s_B \]
Thus, object A will reach a greater height than object B.

### Conclusion
Object A, which has a greater mass and experiences the same air resistance as object B, will go higher than object B when both are thrown upward with the same speed.

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