A) 8.2 m
B) 9.0 m
C) 11.6 m
D) 12.7 m
Correct Answer: A
Solution :
For ball projected with velocity u, it has two components horizontal \[({{\mu }_{x}})\] and vertical\[({{\mu }_{y}})\]. For horizontal distance. \[Time\,(t)=\frac{dis\tan ce\,(d)}{{{u}_{x}}}\] Given, d = 50 m, \[{{\mu }_{x}}=25\,\cos {{60}^{o}}=12.5\,m/s\] \[\therefore \] \[t=\frac{50}{12.5}=4\,s\] For vertical displacement \[y={{u}_{y}}t-\frac{1}{2}g{{t}^{2}}\] Given, \[{{u}_{y}}=u\sin \,{{60}^{o}}\] \[=25\frac{\sqrt{3}}{2}=12.5\sqrt{3}m/s\], \[g=9.8\,m/{{s}^{2}}\], t = 4 s \[\therefore \] \[y=12.5\sqrt{3}\times 4-\frac{1}{2}\times 9.8\times {{(4)}^{2}}\] \[\Rightarrow \] \[y=50\sqrt{3}-78.4\] \[\Rightarrow \] y = 86.6 - 78.4 = 8.2 mYou need to login to perform this action.
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