Question

A football is kicked vertically upwards from the ground and a student gazing out the window sees it moving past her at 5 m/s. The window is 15 m hugh. Air resistance is ignored.
How much time does it take to go from the ground to its highest point?

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{v}=5\mathrm{m}/\mathrm{s} \\ \mathrm{s}=15\mathrm{m} \\ \mathrm{a}=-\mathrm{g} \\ {\mathrm{v}}^2-{\mathrm{u}}^2=2\mathrm{as} \\ {\left(5\right)}^2-{\mathrm{u}}^2=2\left(-\mathrm{g}\right)\times 15 \\ {\mathrm{u}}^2=25+2\times 9.8\times 15 \\ {\mathrm{u}}^2=319 \\ \mathrm{u}=\sqrt{319}=17.86\mathrm{m}/\mathrm{s} \\ \mathrm{At}\mathrm{the}\mathrm{maximum}\mathrm{height},\mathrm{the}\mathrm{new}\mathrm{final}\mathrm{velocity}\mathrm{v}\mathrm{is}\mathrm{zero}. \\ {\mathrm{v}}^2-{\mathrm{u}}^2=2\mathrm{as} \\ 0-{\mathrm{u}}^2=2\left(-\mathrm{g}\right){\mathrm{s}}_{\max } \\ {\mathrm{u}}^2-2{\mathrm{gs}}_{\max }=0 \\ {\mathrm{s}}_{\max }=\frac{{\mathrm{u}}^2}{2\mathrm{g}} \\ {\mathrm{s}}_{\max }=\frac{319}{2\times 9.8} \\ {\mathrm{s}}_{\max }=16.27\mathrm{m} \\ \mathrm{Also},\mathrm{v}=\mathrm{u}+\mathrm{at} \\ \mathrm{At}\mathrm{maximum}\mathrm{height},\mathrm{v}=0 \\ 0=\mathrm{u}-{\mathrm{gt}}_{\max } \\ \mathrm{u}={\mathrm{gt}}_{\max } \\ {\mathrm{t}}_{\max }=\frac{\mathrm{u}}{\mathrm{g}} \\ {\mathrm{t}}_{\max }=\frac{17.86}{9.8} \\ {\mathrm{t}}_{\max }=1.82\mathrm{s} \end{gathered}$$