A) 8.2 m
B) 9.0 m
C) 11.6 m
D) 12.7 m
Correct Answer: A
Solution :
Here ball is hit by cricketer with 25 m/s velocity. The angle of projection \[={{60}^{o}}\] Horizontal component of velocity \[{{\upsilon }_{x}}=25\,\cos {{60}^{o}}\] \[=12.5m/s\] Vertical component of velocity \[{{\upsilon }_{y}}=25\,\,\sin {{60}^{o}}\] \[=25\times \frac{\,\sqrt{3}}{2}=12.5\sqrt{3}m/s\] Time taken by ball to cover 50 m distance \[t=\frac{s}{{{\upsilon }_{x}}}=\frac{50}{12.5}=4\sec \] Now the vertical distance covered by the ball is given by \[x={{\upsilon }_{y}}t-\frac{1}{2}g{{t}^{2}}\] \[=12.5\sqrt{3}\times 4-\frac{1}{2}\times 9.8\times {{(4)}^{2}}\] \[=50\sqrt{3}-78.4\] \[=86.6-78.4\] \[=8.2m\]You need to login to perform this action.
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