Question

A stone is dropped from a height of 100m on earth. At the same time,
another stone is thrown vertically upwards from a ground with a
velocity of 50m/s. At what height from the ground will the stone meat?

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{stone}\mathrm{will}\mathrm{meet}\mathrm{at}\mathrm{time}\mathrm{t}\mathrm{and}\mathrm{at}\mathrm{height}\mathrm{h}, \\ \mathrm{u}=50\mathrm{m}/\mathrm{s} \\ \mathrm{g}=10\mathrm{m}/{\mathrm{s}}^2 \\ \mathrm{height}\mathrm{covered}\mathrm{by}\mathrm{the}\mathrm{stone}\mathrm{thrown}\mathrm{from}\mathrm{ground}, \\ \mathrm{h}=50\mathrm{t}-\frac{1}{2}{\mathrm{gt}}^2 \\ \mathrm{h}=50\mathrm{t}-5{\mathrm{t}}^2.....\left(1\right) \\ \mathrm{distance}\mathrm{covered}\mathrm{by}\mathrm{the}\mathrm{stone}\mathrm{thrown}\mathrm{downward}\mathrm{in}\mathrm{time}\mathrm{t}, \\ \mathrm{since}\mathrm{it}\mathrm{falls}\mathrm{from}\mathrm{rest},\mathrm{u}=0 \\ 100-\mathrm{h}=\frac{1}{2}{\mathrm{gt}}^2 \\ 100-\left(50\mathrm{t}-5{\mathrm{t}}^2\right)=5{\mathrm{t}}^2 \\ 50\mathrm{t}=100 \\ \mathrm{t}=2\mathrm{s} \\ \mathrm{Thus}\mathrm{the}\mathrm{stone}\mathrm{will}\mathrm{meet}\mathrm{each}\mathrm{other}\mathrm{at}2\mathrm{s}. \\ \mathrm{Height}\mathrm{at}\mathrm{which}\mathrm{particle}\mathrm{meet}\mathrm{is}\mathrm{h}. \\ \mathrm{h}=50\mathrm{t}-5{\mathrm{t}}^2 \\ \mathrm{h}=50\left(2\right)-5{\left(2\right)}^2 \\ \mathrm{h}=100-20=80\mathrm{m} \end{gathered}$$