Question

A body $P$ is thrown vertically up with velocity $30m{s}^{-1}$ and another body $Q$ is thrown up along the same vertically line with the same velocity but $1$ second later from the ground. When they meet $\left(g=10m{s}^{-2}\right)$

• A. $P$ travels for $2.5s$
• B. $Q$ travels for $3.5s$
• C. $P$ travels for $3.5s$
• D. $Q$ travels for $1s$

Answer 1

The correct option is C $P$ travels for $3.5s$

$\begin{array}{c}h=ut-\frac{1}{2}g{t}^2\\ h_p=30\times 1-\frac{1}{2}\times 10\times 1^2\\ h_p=30-5=25m\\ v=u-gt\\ v_p=30-10\times 1=20m{s}^{-1}\end{array}$
Let, $t$ further time they take to meet and meet at height $h$
$\begin{array}{l}h = ut - \frac{1}{2}g{t^2}\\h = {h_p} + {u_p}t - \frac{1}{2}g{t^2}\\h = {u_a}t - \frac{1}{2}g{t^2}\\{h_p} + {u_p}t - \frac{1}{2}g{t^2} = {u_a}t - \frac{1}{2}g{t^2}\\{h_p} = \left( {{u_a} - {u_p}} \right)t\\\frac{{{h_p}}}{{{u_a} - {u_p}}} = t\\\frac{{25}}{{30 - 20}} = t\\t = 2.5\,{\rm{secconds}}\\
\end{array}$
Therefore,
$P$ travels $1+t=3.5$