A) 6.0 m
B) 10.0 m
C) 12.0 m
D) 8.0 m
Correct Answer: B
Solution :
For getting horizontal range, there must be some inclination of spring with ground to project ball. Þ \[{{R}_{max}}=\frac{{{u}^{2}}}{g}\] ?..(i) But K.E. acquired by ball = P.E. of spring gun \[\Rightarrow \frac{1}{2}m{{u}^{2}}=\frac{1}{2}k{{x}^{2}}\] Þ \[{{u}^{2}}=\frac{k{{x}^{2}}}{m}\] ?..(ii) From equation (i) and (ii) \[{{R}_{max}}=\frac{k{{x}^{2}}}{mg}\]\[=\frac{600\times {{(5\times {{10}^{-2}})}^{2}}}{15\times {{10}^{-3}}\times 10}\]=10m.You need to login to perform this action.
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