Question

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.

Answer 1

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,

`\begin{gathered} \text{mgh}=(1)/(2)mv^2 \\ gh=(1)/(2)v^2 \end{gathered}`

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

So the height, h will be

`h=(v^2)/(2g)`

Substituting the values, we get

`\begin{gathered} h=((18)^2)/(2*9.8) \\ =16.53\text{ m} \end{gathered}`

So the toadstool is 16.53 m high.