Question

Two particles $1$ and $2$ are thrown in the directions shown in figure simultaneously with velocities $5m/s$ and $20m/s$. Initially, particle $1$ is at height $20m$ from the ground. Taking upwards as the positive direction, the velocity of $2$ with respect to $1$ is $(20+x)m/s$. Find the value of $x$.

Answer 1

The velocity of particle 1 at any time $t$ is given as,

$v_1=-u_1-gt=-5-gt$

The velocity of particle 2 at any time $t$ is given as,

$v_2=u_2-gt=20-gt$

Thus, relative velocity of particle 2 with respect to particle 1 is,

$v_{21}=v_2-v_1=25m/s$

$\therefore x+20=25⟹x=5$