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 \[t=\sqrt{\frac{2H}{g}}=\sqrt{\frac{2\times 10}{10}}=\sqrt{2}s\] \[\therefore \,\,\,\,CB={{\upsilon }_{x}}t'=20\sqrt{2}\] \[\therefore \,\,\,\,\,\,\,AB=20+20\sqrt{2}=20(1+\sqrt{2})m\]You need to login to perform this action.
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