Question

Two bodies $1$ and $2$ are projected simultaneously with velocities $V_1=2\text{m/s}$ and $V_2=4\text{m/s}$ . The body $1$ is projected vertically up from the top of a cliff of height $h=10\text{m}$ and the body $2$ is projected vertically up from the bottom of the cliff . If the bodies meet, find the time in $\left(s\right)$ of meeting of the bodies.

• A. $2\text{s}$
• B. $3\text{s}$
• C. $5\text{s}$
• D. $6\text{s}$

Answer 1

The correct option is C $5\text{s}$

The condition can be shown in the below figure.
Let both the bodies meet at the point $r$


Applying the equation of motion for body $1$ , we get
$S_1=2t-\frac{1}{2}g{t}^2$
Now, applying the equation of motion for body $2$ , we get
$S_2=4t-\frac{1}{2}g{t}^2$
when the two bodies meet, $\Rightarrow 10+S_1=S_2$
$\Rightarrow 10+2t-\frac{1}{2}g{t}^2=4t-\frac{1}{2}g{t}^2$
$\Rightarrow 10+2t=4t$
$\Rightarrow 2t=10$
$\Rightarrow t=5\text{s}$