Question

A body $A$ is dropped from the top of a cliff $42m$ high and at the same time, another body $B$ is thrown vertically upwards from the ground with a velocity of $21m/s$. When and where will they meet?

Answer 1

$\begin{array}{c}x=ut+\frac{1}{2}a{t}^2\\ x=\frac{1}{2}\times g{t}^2......\left(1\right)\\ u_2-x=21t-\frac{1}{2}\times g{t}^2......\left(2\right)\\ u_2-\frac{1}{2}\times g{t}^2=21t-\frac{1}{2}\times g{t}^2\\ t=2\sec \\ x=\frac{1}{2}\times g\times 4=2\times 9.81=18.62\left(fromtop\right)\\ 42-18.62=21.38\left(forbottom\right)\end{array}$