question_answer

A ball is projected from the floor of a long hall having a roof height of\[H=10m\]. The ball is projected with a velocity of \[u=25\text{ }m{{s}^{-1}}\]making an angle of \[\theta =37{}^\circ \] to the horizontal. On hitting the roof the ball loses its entire vertical component of velocity but there is no change in the horizontal component of its velocity. The ball was projected from point A and it hits the floor at B. Find the distance between AB.  

A) \[10(1+\sqrt{2})m\]       

B)        \[15(1+\sqrt{3})m\]

C) \[20(1+\sqrt{2})m\]       

D)        \[8(1+\sqrt{2})m\]

Correct Answer: C

Solution :

[c] \[{{u}_{x}}=u\cos 37{}^\circ =25\times \frac{4}{5}=20m/s\] \[{{u}_{y}}=u\sin 37{}^\circ =10\times \frac{3}{5}=15m/s\] Let the ball hit the roof at time ?t?.             \[H={{u}_{y}}t-\frac{1}{2}g{{t}^{2}}\] \[10=15t-\frac{1}{2}\times 10{{t}^{2}}\Rightarrow {{t}^{2}}-3t+2=0\]\[\Rightarrow \,\,\,t=\frac{3\pm \sqrt{9-8}}{2}=\frac{3+1}{2}\Rightarrow t=1s,2s\] [2 s is unacceptable. Why?] \[\therefore \,\,\,\,\,\,AC=({{u}_{y}})(1s)=20m\] After collision at P, vertical component of velocity is zero. Time of travel from P to B is given by \[t=\sqrt{\frac{2H}{g}}=\sqrt{\frac{2\times 10}{10}}=\sqrt{2}s\] \[\therefore \,\,\,\,CB={{\upsilon }_{x}}t'=20\sqrt{2}\] \[\therefore \,\,\,\,\,\,\,AB=20+20\sqrt{2}=20(1+\sqrt{2})m\]