Q. 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) :
VMMC MedicalVMMC Medical 2005
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 $
