Given,
Height of the fencepost, h=1.9 m
Angle at which the rock was thrown, θ=25°
The height at which the rock was released, a=1 m
The initial speed of the rock, v₀=10 m/s
Referring to the diagram,

On rearranging the above equation,

On substituting the known values,

Therefore the fencepost is at a distance of 1.93 m