A.575 kg smurf is dropped straight down from a toadstool. If the final velocity of the smurf justbefore impact is 18 m/s, how high was the toadstool? Use the Law of Conservation of Energyto solve.

Respuesta :

Given that mass of the object is, m= 575 kg

Also velocity of the object, v= 18 m/s

According to law of conservation of energy,

[tex]\begin{gathered} \text{mgh}=\frac{1}{2}mv^2 \\ gh=\frac{1}{2}v^2 \end{gathered}[/tex]

Here, g is acceleration due to gravity whose value is 9.8 m/s^2.

So the height, h will be

[tex]h=\frac{v^2}{2g}[/tex]

Substituting the values, we get

[tex]\begin{gathered} h=\frac{(18)^2}{2\times9.8} \\ =16.53\text{ m} \end{gathered}[/tex]

So the toadstool is 16.53 m high.