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

Explanation

Here m=10kg, F=10N
the situation is as shown in the figure

Acceleration along vertical direction
`a_Y=F/m=(10N)/(10 kg)=1ms^-2`
Distance travelled by the block in 4s in vertical direction is
`s_y=1/2 a_yt^2=1/2 times (1ms^-2)(4s)^2=8m`
Distance travelled by the block in 4s in horizontal direction is `s_x= (1.5 ms^-1) (4s)=6m`
The displacement of the block at the end of 4s in
`=sqrt(s_X^2+s_y^2)=sqrt((8m)^2+(6m)^2)=10m`