A) 7 m/s
B) 10 m/s
C) 14 m/s
D) 16 m/s
Correct Answer: C
Solution :
At height 10 m, the kinetic energy is equal to potential energy, then \[\frac{1}{2}mv_{0}^{2}=mgh\] \[\frac{1}{2}v_{0}^{2}=9.8\times 10\] \[v_{0}^{2}=2\times 9.8\times 10\] \[{{v}_{0}}=\sqrt{196}\] \[{{v}_{0}}=14\,m\text{/}s\]You need to login to perform this action.
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