question_answer

A ball is thrown vertically upwards. It attains a speed of 10 m/s when it reaches half of its maximum height. What is the maximum height the ball rises?                         [\[\text{g}=\text{1}0\text{m}/{{\text{s}}^{\text{2}}}\]]

Answer:

                Let ?h? be the maximum height the ball can attain. We know, \[g=\frac{4}{3}\pi lGR\] \[\Rightarrow \]\[\frac{{{g}_{1}}}{{{g}_{2}}}=\frac{{{\rho }_{1}}}{{{\rho }_{2}}}\times \frac{{{R}_{1}}}{{{R}_{2}}}=\frac{d}{4d}\times \frac{r}{2r}=\frac{1}{8}\] \[\therefore \] \[g\therefore {{g}_{2}}=1:8\] Considering the motion upto highest point, we have\[{{v}^{2}}={{u}^{2}}+2gh\] Substituting the value of \[{{u}^{2}}\] in the above equation, we get \[0=(100+10h)+2\times (-10)\times h\] \[\Rightarrow 20h=100+10h\Rightarrow 10h=100\] \[\therefore h=10m\]