A small marble is projected with a velocity of `10 m//s` in a direction `45^(@)`from the y-direction on the smooth inclined plane. Calculate the magnitude `v` of its velocity of after `2 s` (Take `g = 10 m//s^(2))`

Options

A. `10sqrt(2)ms^(-1)`
B. `5 ms^(-1)`
C. `10ms^(-1)`
D. `5sqrt(2)ms^(-1)`

Answer

C

Explanation

Given `(V=10ms^(-1))`
After `2s: V_(x)=(10)/(sqrt(2))-(10)/(sqrt(2)) xx 2 implies V_(x)=-(10)/(sqrt(2)) ms^(-1)`
`V_(x)=-(10)/(sqrt(2)) ms^(-1) and V_(y)=-(10)/(sqrt(2)) ms^(-1)`
`V=sqrt((100)/(2)+(100)/(2))=10sqrt((1)/(2)+(1)/(2))=10ms^(-1)`.