Question

A stone is thrown in a vertically up ward direction with a velocity of 5 ms. if the acceleration of the stone driving its motion is 10 meterper second in the dawn word direction. what will be the eight attained by the stone and how much time will it take to reach there?

Answer 1

$$\begin{gathered} \mathrm{u}=5\mathrm{m}/\mathrm{s} \\ \mathrm{As}\mathrm{the}\mathrm{acceleration}\mathrm{acts}\mathrm{in}\mathrm{the}\mathrm{opposite}\mathrm{direction}\mathrm{of}\mathrm{initial}\mathrm{velocity},\mathrm{so}\mathrm{it}\mathrm{is}\mathrm{negative}. \\ \mathrm{a}=-10\mathrm{m}/{\mathrm{s}}^2 \\ \mathrm{At}\mathrm{the}\mathrm{height}\mathrm{point},\mathrm{v}=0 \\ \mathrm{Using},{\mathrm{v}}^2-{\mathrm{u}}^2=2\mathrm{as} \\ {0}^2-5^2=2\times \left(-10\right)\times \mathrm{s} \\ -25=-20\mathrm{s} \\ \mathrm{s}=\frac{25}{20}=1.25\mathrm{m} \\ \mathrm{Again},\mathrm{v}=\mathrm{u}+\mathrm{at} \\ 0-5=\left(-10\right)\mathrm{t} \\ \mathrm{t}=\frac{5}{10}=0.5\mathrm{s} \\ \mathrm{Time}\mathrm{taken}\mathrm{by}\mathrm{the}\mathrm{stone}\mathrm{to}\mathrm{reach}\mathrm{the}\mathrm{heighest}\mathrm{point}=0.5\mathrm{s} \end{gathered}$$