Question

an engineer is trying to determine the height from which she needs to release a 55kg wrecking ball such that the wrecking ball will be traveling 10.7m/s when it strikes the building at the position shown. neglect drag. list the energy types at the initial and final time and whether work and loss (due to non - conservative forces) occur as well as the corresponding amounts of energy. include the earth as part of the system. initial: = unit work? = unit loss? = unit final: = unit from what height above the ground does the engineer need to release the ball?

Explanation:

Step1: Identify Energy Types

The system includes the ball and Earth, so we use conservation of mechanical energy (since drag is neglected, non - conservative work \(W_{nc}=0\)). Initially, the ball has gravitational potential energy (\(U_{g,i}\)) and kinetic energy (\(K_i = 0\) as it's released from rest). Finally, it has kinetic energy (\(K_f\)) and gravitational potential energy (\(U_{g,f}\)) at height \(h_f = 5.7\space m\) above the ground.

The formula for gravitational potential energy is \(U_g=mgh\) and for kinetic energy is \(K=\frac{1}{2}mv^2\). By conservation of energy (\(E_i = E_f\) because \(W_{nc}=0\)):

$$mgh_i+\frac{1}{2}mv_i^2=mgh_f+\frac{1}{2}mv_f^2$$

Since \(v_i = 0\) (released from rest), the equation simplifies to:

$$mgh_i=mgh_f+\frac{1}{2}mv_f^2$$

We can divide both sides by \(m\):

$$gh_i=gh_f+\frac{1}{2}v_f^2$$

Step2: Solve for Initial Height \(h_i\)

We know that \(g = 9.8\space m/s^2\), \(h_f=5.7\space m\), and \(v_f = 10.7\space m/s\). Rearranging the formula for \(h_i\):

$$h_i=h_f+\frac{v_f^2}{2g}$$

Substitute the values:
First, calculate \(\frac{v_f^2}{2g}=\frac{(10.7)^2}{2\times9.8}=\frac{114.49}{19.6}\approx5.84\space m\)
Then, \(h_i=5.7 + 5.84=11.54\space m\)

Answer:

The engineer needs to release the ball from a height of approximately \(\boldsymbol{11.5\space m}\) (or more precisely \(11.54\space m\)) above the ground.