A) 28m/s
B) 10m/s
C) 14m/s
D) 20m/s
Correct Answer: D
Solution :
[d] When it hits the ground \[TE=\frac{1}{2}mv_{0}^{2}+mgh\] It loses 50% energy, it left with 50% energy, this left energy converted into potential energy \[\frac{50}{100}\left[ \frac{1}{2}mv_{0}^{2}+mgh \right]=mgh\] \[mv_{0}^{2}+2\,mgh=4\,mgh\,\] \[v_{0}^{2}+2\,gh\] \[v_{0}^{2}+2\times 10\times 20\] \[v_{0}^{2}=400\] \[v_{0}^{2}=20m/s\]You need to login to perform this action.
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