Question

A stone is thrown in a vertically upward direction with a velocity of 5 m/s. If the acceleration of the stone during its motion is 10 m/s in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

Answer 1

Dear Student,

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

Regards