Question

a stone thrown upwards from ground level has its equation of height h=490t-4.9t² where 'h' is in metres and 't' is in seconds . what is the maximum height reached by it?

Answer 1

$$\begin{gathered} \mathrm{Given}\mathrm{equation}: \\ \mathrm{h}=490\mathrm{t}-4.9{\mathrm{t}}^2 \\ \mathrm{h}=490\mathrm{t}-\frac{1}{2}\left[9.8\right]{\mathrm{t}}^2 \\ \mathrm{h}=490\mathrm{t}-\frac{1}{2}{\mathrm{gt}}^2 \\ \mathrm{Compare}\mathrm{with}\mathrm{the}\mathrm{standard}\mathrm{equation}: \\ \mathrm{h}=\mathrm{ut}-\frac{1}{2}{\mathrm{gt}}^2 \\ \mathrm{Hence}, \\ \mathrm{Initial}\mathrm{velocity}\mathrm{of}\mathrm{projection},\mathrm{u}=490\mathrm{m}/\mathrm{s} \\ \mathbf{Maximum}\mathbf{}\mathbf{height}\mathbf{}\mathbf{reached}\mathbf{:} \\ \mathrm{h}=\frac{{\mathrm{u}}^2}{2\mathrm{g}} \\ =\frac{{490}^2}{2\times 9.8} \\ =12250\mathrm{m} \end{gathered}$$