Question

If \( V_1 \) is the velocity of a body projected from point A and \( V_2 \) is the velocity of a body projected from point B, which is vertically below the highest point C, and if both the bodies collide, then: 
 

• A. \( V_1 = \frac{1}{2} V_2 \)
• B. \( V_2 = \frac{1}{2} V_1 \)
• C. \( V_1 = V_2 \)
• D. \( V_1 = 3V_2 \)

Hint:

For projectile motion:
- Velocity is affected by gravitational potential energy changes.
- Use energy conservation or kinematic equations to compare speeds.
- Symmetry of projectile paths helps in velocity comparison.

Answer 1

The Correct Option is B

To analyze the motion of the two bodies and determine their velocity relationship at the point of collision, we apply the kinematic equations for projectile motion.
Step 1: Understanding the Given Motion
- The body at point \( A \) is projected with velocity \( V_1 \) at an angle \( \theta \).
- The body at point \( B \) is projected vertically with velocity \( V_2 \).
- The highest point \( C \) represents the peak of the trajectory of the first body.
- Both bodies collide at a certain point, meaning they must meet at the same height and at the same time.
Step 2: Motion of the Body from \( A \)
- The body projected from \( A \) reaches the highest point \( C \) with a vertical velocity component of zero.
- The initial vertical velocity of the body from \( A \) is:
\[ u_{Ay} = V_1 \sin \theta \] - The time taken to reach the highest point \( C \) is given by:
\[ t_C = \frac{u_{Ay}}{g} = \frac{V_1 \sin \theta}{g} \] - The total height of \( C \), using the equation of motion \( v^2 = u^2 + 2as \), is:
\[ h = \frac{(V_1 \sin \theta)^2}{2g} \] Step 3: Motion of the Body from \( B \)
- The second body at \( B \) is projected vertically upwards with velocity \( V_2 \).
- Let the time taken by the second body to reach the point of collision be \( t \).
- The equation of motion for vertical displacement is:
\[ h' = V_2 t - \frac{1}{2} g t^2 \] Since both bodies collide at the same height and time, setting \( h = h' \), we equate the expressions:
\[ \frac{(V_1 \sin \theta)^2}{2g} = V_2 t - \frac{1}{2} g t^2 \] Substituting \( t = \frac{V_1 \sin \theta}{g} \), we get:
\[ \frac{(V_1 \sin \theta)^2}{2g} = V_2 \times \frac{V_1 \sin \theta}{g} - \frac{1}{2} g \left(\frac{V_1 \sin \theta}{g}\right)^2 \] Simplifying,
\[ \frac{(V_1 \sin \theta)^2}{2g} = \frac{V_1 V_2 \sin \theta}{g} - \frac{(V_1 \sin \theta)^2}{2g} \] Rearranging for \( V_2 \),
\[ V_2 = \frac{1}{2} V_1 \] Thus, the correct answer is:
\[ {V_2 = \frac{1}{2} V_1} \]