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

Answer:

164.29 m

Explanation:

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

$\begin{gathered} E_i=E_f \\ \text{mgh}+\frac{1}{2}mv^2_i=\frac{1}{2}mv^2_f \end{gathered}$

Solving for the height h, we get:

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

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:

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

Therefore, the height of the building is 164.29 m