A) \[14.62\,m/s,60{}^\circ \]
B) \[14.62\,m/s,ta{{n}^{-1}}\]
C) \[22.36\,m/s,ta{{n}^{-1}}\]
D) \[22.36\,m/s,\,60{}^\circ \]
Correct Answer: C
Solution :
[c] Total time of flight is \[T=4\]s and if u is its initial speed and \[\theta \]the angle of projection. Then \[T=\frac{2u\sin \theta }{g}=4\] Or \[u\,\sin \theta =2g\] ...(1) After 1 second velocity vector makes an angle of \[45{}^\circ \] with horizontal i.e. \[{{v}_{x}}={{v}_{y}}\] or \[u\cos \theta =(u\sin \theta )-(gt)\] or \[u\cos \theta =2g-g\] \[(t=1)\] or \[u\cos \theta =g\] ...(2) Squaring and adding (1) and (2), we get \[{{u}^{2}}=5{{g}^{2}}=5{{(10)}^{2}}{{m}^{2}}/{{s}^{2}}\] \[\therefore \,\,\,\,\,\,\,\,u=22.36\,m/s\] Dividing (1) and (2) we get, \[\tan \theta =2\] or \[\theta ={{\tan }^{-1}}(2)\]
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