1 \(10\sqrt 3 \,m{s^{ - 1}}\)
2 \(20\sqrt 3 \,m{s^{ - 1}}\)
3 \(10\sqrt 5 \,m{s^{ - 1}}\)
4 \(20\sqrt 2 \,m{s^{ - 1}}\)
Explanation:
Suppose the angle made by the instantaneous velocity with the horizontal be \(\alpha \). Then
\({\rm{tan}}\,{\rm{\alpha = }}\frac{{{v_y}}}{{{v_x}}} = \frac{{u\sin \theta - gt}}{{u\cos \theta }}\)
Given that \(\alpha = 45^\circ ,\,{\rm{when}}\,t = 1\,s;\alpha = 0^\circ \), when \(t = 2s\)
This gives \(u\cos \theta = u\sin \theta - g\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)\)
and \(u\sin \theta - 2g = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)\)
From eqs. (1) and (2), we find \(u\sin \theta = 2g\) and \(u\cos \theta = g\).
Squaring and adding, \(u = \sqrt 5 g = 10\sqrt 5 {\mkern 1mu} {\mkern 1mu} m{s^{ - 1}}\)