question_answer

A stone is thrown vertically upwards. When the particle is at a height half of the maximum height, its speed is 10 m/s, then find maximum height attained by particle. \[(g=10\text{ }m{{s}^{-2}})\]

Answer:

                Consider the maximum height attained is ?h?. \[g=10m/{{s}^{2}}\] At the half of the the maximum height. \[u=10m/s.\] At maximum height\[v=0\] Using \[{{v}^{2}}{{u}^{2}}=2gh\] \[\Rightarrow \] \[{{0}^{2}}{{(10)}^{2}}=2\times (-10)\times \frac{h}{2}\]\[\Rightarrow \]\[\frac{h}{2}=\frac{{{10}^{2}}}{20}\] \[\frac{h}{2}=\frac{100}{20}=5\] \[\therefore \]\[{{h}_{\max }}=10m\]