Question

A PARTICLE IS PROJECTED VERTICALLY UPWARDS FROM A POINT 'A' ON THE GROUND .IT TAKES TIME T₁TO REACH POINT 'B' BUT IT STILL CONTINUES TO MVE UP .IF IT TAKES TIME T₂ TO REACH THE GROUND FROM POINT B .THEN THE HEIGHT OF POINT B FROM THE GROUND IS ?

Answer 1

$$\begin{gathered} \mathrm{Let} \\ \mathrm{Height}\mathrm{of}\mathrm{point}\mathrm{B}=\mathrm{h} \\ \mathrm{Total}\mathrm{time}\mathrm{of}\mathrm{flight}=\mathrm{T}1+\mathrm{T}2 \\ \mathrm{Time}\mathrm{taken}\mathrm{to}\mathrm{reach}\mathrm{maximum}\mathrm{height},\mathrm{t}=\frac{\mathrm{T}1+\mathrm{T}2}{2} \\ \mathrm{Velocity}\mathrm{of}\mathrm{projection}=\mathrm{u} \\ \mathbf{Motion}\mathbf{}\mathbf{from}\mathbf{}\mathbf{ground}\mathbf{}\mathbf{to}\mathbf{}\mathbf{maximum}\mathbf{}\mathbf{height}\mathbf{:} \\ \mathrm{Final}\mathrm{velocity}\mathrm{at}\mathrm{maximum}\mathrm{height}\mathrm{is}, \\ \mathrm{v}=\mathrm{u}-\mathrm{gt} \\ \mathrm{At}\max \mathrm{height},\mathrm{final}\mathrm{velocity}\mathrm{is}\mathrm{zero}. \\ \mathrm{Hence}, \\ 0=\mathrm{u}-\mathrm{g}\left(\frac{\mathrm{T}1+\mathrm{T}2}{2}\right) \\ \mathrm{u}=\mathrm{g}\left(\frac{\mathrm{T}1+\mathrm{T}2}{2}\right) \\ \mathbf{Motion}\mathbf{}\mathbf{from}\mathbf{}\mathbf{ground}\mathbf{}\mathbf{to}\mathbf{}\mathbf{point}\mathbf{}\mathbf{B}\mathbf{:} \\ \mathrm{Displacement}\mathrm{is}, \\ \mathrm{s}=\mathrm{u}\left(\mathrm{T}1\right)-\frac{1}{2}\mathrm{g}{\left(\mathrm{T}1\right)}^2 \\ \mathrm{h}=\mathrm{g}\left(\frac{\mathrm{T}1+\mathrm{T}2}{2}\right)\left(\mathrm{T}1\right)-\frac{1}{2}\mathrm{g}{\left(\mathrm{T}1\right)}^2 \\ =\frac{1}{2}\mathrm{g}{\left(\mathrm{T}1\right)}^2+\frac{1}{2}\mathrm{g}\left(\mathrm{T}1\right)\left(\mathrm{T}2\right)-\frac{1}{2}\mathrm{g}{\left(\mathrm{T}1\right)}^2 \\ =\frac{\mathbf{1}}{\mathbf{2}}\mathbf{g}\left(\mathbf{T}\mathbf{1}\right)\left(\mathbf{T}\mathbf{2}\right) \end{gathered}$$