A) 490
B) 98
C) 49
D) 0.5
Correct Answer: C
Solution :
Vertical velocity after 10 s is \[u=(98\,sin{{60}^{o}})-9.8\left( \frac{\sqrt{3}}{2}-1 \right)\] \[=98(0.866-1)=-98\times 0.134\,m/s\] Vertical momentum of the ball after 10 s = mv \[=0.5\times (-98\times 0.134)kg.m/s\] Initial vertical momentum of the ball \[=mv\sin {{60}^{o}}\] \[=0.5\times 98\times \frac{\sqrt{3}}{2}=0.5\times 98\times 0.866\] Change in vertical momentum \[=0.5\times 98\times 0.866-(-0.5\times 98\times 0.134)\] \[=0.5\times 98[0.866+0.134]\] \[=0.5\times 98\] \[=49\,kg.m/s\]
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