question_answer

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