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 :

                 For ball projected with velocity u, it has two components horizontal \[({{\mu }_{x}})\] and vertical\[({{\mu }_{y}})\].  For horizontal distance.                                 \[Time\,(t)=\frac{dis\tan ce\,(d)}{{{u}_{x}}}\] Given,   d = 50 m,                 \[{{\mu }_{x}}=25\,\cos {{60}^{o}}=12.5\,m/s\] \[\therefore \]  \[t=\frac{50}{12.5}=4\,s\] For vertical displacement                 \[y={{u}_{y}}t-\frac{1}{2}g{{t}^{2}}\] Given,   \[{{u}_{y}}=u\sin \,{{60}^{o}}\]                 \[=25\frac{\sqrt{3}}{2}=12.5\sqrt{3}m/s\],                 \[g=9.8\,m/{{s}^{2}}\],  t = 4 s \[\therefore \]  \[y=12.5\sqrt{3}\times 4-\frac{1}{2}\times 9.8\times {{(4)}^{2}}\] \[\Rightarrow \]               \[y=50\sqrt{3}-78.4\] \[\Rightarrow \]               y = 86.6 - 78.4 = 8.2 m