question_answer

If a body is projected vertically up, its velocity decreases to half of its initial velocity at a height 'h' above the ground. What is the maximum height reached by it?

Answer:

For AB Initial velocity = u Final velocity \[=(v)\frac{u}{{}}\] We know that, \[{{v}^{2}}-{{u}^{2}}=-2gh\] \[\Rightarrow {{\left( \frac{u}{2} \right)}^{2}}-{{u}^{2}}=-2gh\] \[\Rightarrow \frac{{{u}^{2}}-4{{u}^{2}}}{4}=-2gh\Rightarrow \frac{{{u}^{2}}}{g}=\frac{8h}{3}\]                  ?..(1) Maximum height \[h=\frac{{{u}^{2}}}{2g}\]                         ?..(2) Substituting (1) in (2), we get \[=\frac{1}{2}\left[ \frac{{{u}^{2}}}{g} \right]=\frac{1}{2}\times \frac{8h}{3}=\frac{4h}{3}\]