Question

What is the height of a building if Kevin can throw a ball down with an initial velocity of 12m/s and it lands with a final velocity of 58m/s.

Answer 1

164.29 m

Step-by-step explanation:

By the conservation of energy, we can write the following equation:

`\begin{gathered} E_i=E_f \\ \text{mgh}+(1)/(2)mv^2_i=(1)/(2)mv^2_f \end{gathered}`

Solving for the height h, we get:

`\begin{gathered} \text{mgh}=(1)/(2)mv^2_f-(1)/(2)mv^2_i \\ \text{mgh}=(1)/(2)m(v^2_f-v^2_i) \\ h=(1)/(2mg)m(v^2_f-v^2_i) \\ h=(1)/(2g)(v^2_f-v^2_i) \end{gathered}`

So, replacing the initial velocity vi = 12 m/s, the final velocity vf = 58 m/s and the gravity g = 9.8 m/s², we get:

`\begin{gathered} h=(1)/(2(9.8))(58^2-12^2) \\ h=164.29\text{ m} \end{gathered}`

Therefore, the height of the building is 164.29 m