A block of mass 10 kg is moving horizontally with a speed of `1.5ms^(-1)` on a smooth plane. If a constant vertical force 10 N acts on it, the displacement of the block from the point of application of the force at the end of 4 s is

Options

A. 5 m
B. 20 m
C. 12 m
D. 10 m

Answer

D

Explanation

`S_("horizontal")=ut=1.5xx4=6m`
`S_("vertical")=(1)/(2)at^(2)=(1)/(2)(F)/(m)t^(2)`
`=(1)/(2)xx(10)/(10)xx(4)^(2)=8m`
`S_("net")=sqrt((6)^(2)+(8)^(2))`
`=sqrt(36+64)=sqrt(100)=10m`