Question

From the top of tower a body $A$ is projected vertically up, another body $B$ is horizontally thrown and a third body $C$ is thrown vertically down with the same velocity. Then,

• A. $B$ strikes the ground with more velocity
• B. $C$strikes the ground with less velocity
• C. $A,BandC$ strike the ground with same velocity
• D. $AandC$ strike the ground with more velocity than $B$

Answer 1

The correct option is C

$A,BandC$ strike the ground with same velocity

Step 1: Given data:

$A$ is projected vertically up, another body $B$ is horizontally thrown and a third body $C$ is thrown vertically down with the same velocity.

The diagram represents the trajectory followed by three particles.

Step 2: For $A:$ It goes with velocity, $u$ will reach its maximum height (velocity becomes zero), and comes back and attain a velocity $u$

Using the equation of motion,

$$\begin{gathered} v_A^2=u^2+2gh \\ {v}_A=\sqrt{u^2+2gh} \end{gathered}$$ [ where, $g$is the acceleration due to gravity, and $h$is the maximum height]

Step 3: For $B:$ Horizontal velocity $v_x=u$ remains the same

Vertical velocity,

$$\begin{gathered} v_y=\sqrt{0+2gh} \\ {v}_y=\sqrt{2gh} \end{gathered}$$

The resultant velocity,

$$\begin{gathered} v_B=\sqrt{v_x^2+v_y^2} \\ {v}_B=\sqrt{u^2+2gh} \end{gathered}$$

Step 4: For $C:$ Going down with velocity $u$

$v_C=\sqrt{u^2+2gh}$

If the force or acceleration acting on various bodies having the same initial velocity is the same, the final velocity will also be the same.

All three particles have the same initial velocity and the only force acting on them is gravity. So, the final velocity will be the same for all particles no matter what trajectory they followed.

Therefore, A, B, and C strikes the ground with the same velocity.

Hence, option C is correct.