Question

a stone is dropped from the top of a tall cliff and 1 sec later a second stone is throw vertically downward with a velocity of 20 m per sec . how far below the top of the cliff will the second stone overtake the first

Answer 1

$$\begin{gathered} Forthefirststone: \\ intialspeedu=0 \\ dis\tan cetravelledintseconds \\ s=ut+\frac{1}{2}a{t}^2 \\ s=\frac{1}{2}g{t}^2 \\ Secondstoneisthrownafteronesecondwithintialvelocityu=20m/s,thereforesamedistnaceStravelledbysecondstoneintimet-1is: \\ s=20(t-1)+\frac{1}{2}g(t-1{)}^2 \\ fromtheseequationswegett=2.5s \\ s=\frac{1}{2}\times 10\times 2.5\times 2.5 \\ s=125m \\ after125msecondstonewillovertakefirststone. \end{gathered}$$