A) 10 m/s
B) \[10\sqrt{30}\]m/s
C) 40 m/s
D) 20 m/s
Correct Answer: C
Solution :
Ball starts from the top of a hill which is 100 m high and finally rolls down to a horizontal base which is 20 m above the ground so from the conservation of energy \[mg\,({{h}_{1}}-{{h}_{2}})=\frac{1}{2}m{{v}^{2}}\] Þ \[v=\sqrt{2g({{h}_{1}}-{{h}_{2}})}=\sqrt{2\times 10\times (100-20)}\] \[=\sqrt{1600}=40\,m/s\].You need to login to perform this action.
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