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A can sits on a vertical wooden fencepost 1.9 meters above the ground. Billy picks up a small rock, aims at an angle ϴ = 25⁰ above the horizontal and throws the rock, releasing it 1 m above the ground with an initial speed of v0 =10 m/s. Boom! He hits the can! How far away is the fencepost?

Sagot :

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

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