A) \[2.86\,\,m\]
B) \[1.88\,\,m\]
C) \[3.82\,\,m\]
D) \[1.84\,\,m\]
Correct Answer: D
Solution :
\[K.E.=\frac{1}{2}m{{v}^{2}}\] \[=\frac{1}{2}(20){{(6)}^{2}}=\frac{1}{2}\times 20\times 36\,\,J=360\,\,J\] At the highest point, the kinetic energy becomes zero, thus the entire \[K.E.\] of \[360\,\,J\] is converted into\[P.E\]. So the \[P.E.\] at the highest point is\[360\,\,J\]. \[P.E.=360=20\times 9.8\times h\] \[h=\frac{360}{20\times 9.8}=1.84\,\,m\] \[h=\frac{360}{20\times 9.8}=1.84\,\,m\]
You need to login to perform this action.
You will be redirected in
3 sec
