question_answer

A cricketer hits a ball with a velocity \[25\,\,m/s\] at \[{{60}^{o}}\] 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) [BVP 2003]

A)             8.2 m  

B)             9.0 m

C)             11.6 m

D)               12.7 m

Correct Answer: A

Solution :

                Horizontal component of velocity                   \[{{v}_{x}}=25\cos 60{}^\circ =12.5\,m/s\]             Vertical component of velocity                   \[{{v}_{y}}=25\sin 60{}^\circ =12.5\sqrt{3}\,m/s\]             Time to cover 50 m distance \[t=\frac{50}{12.5}=4\,\sec \]             The vertical height y is given by             \[y={{v}_{y}}t-\frac{1}{2}g{{t}^{2}}=12.5\sqrt{3}\times 4-\frac{1}{2}\times 9.8\times 16=8.2\,m\]