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,
[tex]\tan \theta=\frac{h-a}{d}[/tex]On rearranging the above equation,
[tex]d=\frac{h-a}{\tan \theta}[/tex]On substituting the known values,
[tex]d=\frac{1.9-1}{\tan 25^0}=1.93\text{ m}[/tex]Therefore the fencepost is at a distance of 1.93 m
