Question

Two particles A and B are projected from same point with same speed at same time as shown in the figure:

Particle A is projected down and B is projected up. For this situation, mark out the correct statements

• A. Both reach the ground at same time
• B. Both reach the ground at different time
• C. Both reach the ground with same speed
• D. Both have same horizontal displacement when they reach the ground.

Answer 1

Answer: (B), (C)

Both reach the ground with same speed

Explanation for correct options:
Option B:

1. The vertical component of velocity of both the particles:
   $$\begin{gathered} u_{Ay}=u_0\sin \theta \\ {u}_{By}=-u_0\sin \theta \end{gathered}$$
2. Let $t_A\&t_B$ represent the time taken by the particles A and B to reach the ground.
3. Use the second equation of motion on both these particles in the vertical direction:
   $$\begin{gathered} h=u_{Ay}{t}_A-\frac{1}{2}g{t_A}^2=u_0\sin \theta \times t_A-\frac{1}{2}g{t_A}^2 \\ h=u_{By}{t}_B-\frac{1}{2}g{t_B}^2=-u_0\sin \theta \times t_B-\frac{1}{2}g{t_B}^2 \end{gathered}$$

Clearly from the above two quadratic equations, the value of $t_A\&t_B$ will be different.
Option C:

1. Since there is no horizontal acceleration, the horizontal component of final velocity remains the same as the horizontal component of initial velocity:
   $$\begin{gathered} v_{Ax}=u_{Ax}=u_0\cos \theta \\ {v}_{Bx}=u_{Bx}=u_0\cos \theta \end{gathered}$$
2. Use the third equation of motion on both these particles in the vertical direction:
   $$\begin{gathered} {v_{Ay}}^2={u_{Ay}}^2-2gh={u_0}^2{\sin }^2\theta -2gh \\ {v_{By}}^2={u_{By}}^2-2gh={u_0}^2{\sin }^2\theta -2gh \end{gathered}$$
3. So, the horizontal and vertical components of the final velocities of particle A are equal to that of particle B.
4. Thus, their final velocities will be equal.

Clearly from the above equations, the final speed of both the particles will be equal when they reach the ground.

Explanation for incorrect options:
Option A:

1. Particle B first goes up, then comes to the same level from which it started.
2. Then it moves down and covers the same vertical distance as A.
3. Clearly, the vertical distance traveled by the particle B is more than particle A.
4. Since the speed was the same, it will take more time for B than A.

Hence, both particles cannot reach the ground at the same time.

Option D:

1. Particle B takes more time than A to reach the ground.
2. The horizontal component of velocities of both the particles remains the same as there is no change in horizontal acceleration.
3. Due to more time but the same speed, particle B covers more horizontal distance than A.

Hence, both particles do not have the same horizontal displacement when they reach the ground.
Thus, both the particles will reach the ground at different times, but with the same speed. So options B and C are correct.