Answer:
13.2 m
Note: This answer was rounded to three significant figures.
Explanation:
To determine the height from which the engineer needs to release the wrecking ball so that it reaches a speed of 11.7 m/s when it strikes the building, we can use the principles of conservation of energy.
[tex]\boxed{ \begin{array}{ccc} \text{\underline{Energy Conservation:}} \\\\ \text{Total Energy at Start} = \text{Total Energy at End} \\\\ E_{\text{total, initial}} = E_{\text{total, final}} \\\\ \text{Where:} \\ \bullet \ E_{\text{total, initial}} \ \text{is the total initial energy (kinetic + potential + others)} \\ \bullet \ E_{\text{total, final}} \ \text{is the total final energy (kinetic + potential + others)} \end{array} }[/tex]
[tex]\boxed{\begin{array}{ccc}\text{\underline{Gravitational Potential Energy: }}\\\\U_g=mgy\\\\\text{Where:}\\\bullet \ U_g\ \text{is the measure of potential energy}\\\bullet \ m\ \text{is the mass of the object}\\\bullet \ g\ \text{is the acceleration due to gravity} \ (9.8 \ m/s^2)\\\bullet \ y\ \text{is the height above a reference point}\end{array}}[/tex]
[tex]\boxed{ \begin{array}{ccc} \text{\underline{Kinetic Energy:}} \\\\ K = \dfrac{1}{2}mv^2 \\\\ \text{Where:} \\ \bullet \ K \ \text{is the kinetic energy} \\ \bullet \ m \ \text{is the mass of the object} \\ \bullet \ v \ \text{is the velocity of the object} \end{array}}[/tex]
Energy Analysis
The initial energy of the system (E₀):
- Potential Energy: The ball has gravitational potential energy at the height 'h'
- Kinetic Energy: The ball is released from rest, so its initial kinetic energy is zero
Thus,
[tex]\Longrightarrow E_0=mgh[/tex]
[tex]\Longrightarrow E_0=(69 \text{ kg})(9.8 \text{ m/s}^2)h[/tex]
[tex]\therefore E_0=(666.4 \text{ N})h[/tex]
The final energy of the system (E_f):
- Potential Energy: The potential energy when the ball is 6 meters above the reference level (ground)
- Kinetic Energy: The ball is moving with a velocity of 11.7 m/s just before impact
Thus,
[tex]\Longrightarrow E_f=mgy+\dfrac{1}{2}mv^2[/tex]
[tex]\Longrightarrow E_f=(69 \text{ kg})(9.8 \text{ m/s}^2)(6 \text{ m})+\dfrac{1}{2}(69 \text{ kg})(11.7 \text{ m/s})^2[/tex]
[tex]\therefore E_f=8779.91 \text{ J}[/tex]
Using energy conservation:
[tex]E_0=E_f[/tex]
[tex]\Longrightarrow (666.4 \text{ N})h=8779.91 \text{ J}[/tex]
Solving for 'h':
[tex]\Longrightarrow h=\dfrac{8779.91 \text{ J}}{666.4 \text{ N}}\\\\\\\\ \therefore h \approx \boxed{13.2 \text{ m}}[/tex]
Thus, the engineer needs to release the wrecking ball from a height of approximately 13.2 meters above the ground for it to reach a speed of 11.7 m/s when it strikes the building.
