Question

A stone is thrown vertically upwards with an initial velocity of 14m/s. Find the maximum height reached and the time of ascent

Answer 1

$$\begin{gathered} \mathbf{Max}\mathbf{}\mathbf{height}\mathbf{:} \\ \mathrm{H}=\frac{{\mathrm{u}}^2}{2\mathrm{g}} \\ =\frac{{14}^2}{2\times 9.8} \\ =10\mathrm{m} \\ \mathbf{Time}\mathbf{}\mathbf{of}\mathbf{}\mathbf{ascent}\mathbf{:} \\ \mathrm{T}=\frac{\mathrm{u}}{\mathrm{g}} \\ =\frac{14}{9.8} \\ =1.4\mathrm{s} \end{gathered}$$