question_answer

A cricketer hits a ball with a velocity 25 m/s at \[60{}^\circ \] above the horizontal. How far above the ground it passes over a fielder 50 m from the bat (assume the ball is struck very close to the ground) :

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 \[{{\upsilon }_{x}}=25\cos {{60}^{o}}\] \[=12.5\,m/s\] 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 \[x={{\upsilon }_{y}}t-\frac{1}{2}g{{t}^{2}}\] \[=12.5\sqrt{3}\times 4-\frac{1}{2}\times 9.8\times {{(4)}^{2}}\] \[=50\sqrt{3}-78.4\] \[86.6-78.4\] \[=8.2\,m\]