Question

A stone is thrown vertically upwards with an initial velocity of14m/s. Find maximum ht reached and time of ascent (g=9.8m/s2)

Answer 1

$$\begin{gathered} Given, \\ \mathrm{Initial}\mathrm{velocity}\mathrm{of}\mathrm{ball},u=14\mathrm{m}{\mathrm{s}}^{-1} \\ \mathrm{Maximum}\mathrm{height}\mathrm{reached}\left(H\right)\mathrm{is}\mathrm{given}\mathrm{by} \\ H=\frac{u^2{\sin }^2\theta }{2g}=\frac{{\left(14\right)}^2{\sin }^{2}90°}{2\times 9.8} \\ \Rightarrow H=\frac{196}{19.6}=10\mathrm{m} \\ \mathrm{Time}\mathrm{of}\mathrm{ascent}\left(t\right)\mathrm{is}\mathrm{given}\mathrm{by} \\ t=\frac{u\sin \theta }{g}=\frac{14\sin 90°}{9.8} \\ \Rightarrow t=1.42\mathrm{s} \end{gathered}$$