Physics For Scientists And Engineers: A Strategic Approach, 4th Edition
Authors: Randall D. Knight
ISBN-13: 978-0133942651
See our solution for Question 13E from Chapter 10 from Randall Knight's Physics for Scientists and Engineers.
Problem 13E
Chapter:
Problem:
A cannon tilted up at a 30° angle fires a cannon ball at 80 m/s from atop a 10-m-high fortress wall. What is the ball's impact speed on the ground below? Ignore air resistance.
Step-by-Step Solution
Step 1
We will use the conservation of energy to find the speed of ball on impact.
Step 2: Initial Energy of ball,
Kinetic energy of ball (mass $m$ and initial velocity $v_1$) before firing. \[{K_1} = \dfrac{1}{2}mv_1^2\]Potential energy of ball mass $m$ and initial height $y_1$) before firing\[{U_1} = mg{y_1}\]
Step 3: Final Energy of ball,
Kinetic energy of ball (mass $m$ and final velocity $v_2$) upon impact \[{K_2} = \dfrac{1}{2}mv_2^2\]Potential energy of ball (mass $m$ and final height $y_2$) upon impact with mass $m$ and initial height velocity $y_1$. \[{U_2} = mg{y_2}\]
Step 4: Applying conservation of Total energy
Energy before firing is equal to energy after impact\[\begin{array}{l}{K_1} + {U_1} = {K_2} + {U_2}\\\\\dfrac{1}{2}mv_1^2 + mg{y_1} = \dfrac{1}{2}mv_2^2 + mg{y_2}\end{array}\]
Step 5: Final velocity
Initial height of ball is 10 m. Upon impact height becomes 0. Intital velocity is given by 80 m/s.\[\begin{array}{l}\dfrac{1}{2}mv_1^2 + mg{y_1} = \dfrac{1}{2}mv_2^2 + mg{y_2}\\\\\dfrac{1}{2}v_1^2 + g{y_1} = \dfrac{1}{2}v_2^2 + g{y_2}\\\\\dfrac{1}{2}{\left( {80} \right)^2} + 9.8\left( {10} \right) = \dfrac{1}{2}v_2^2 + g\left( 0 \right)\\\\{v_2} = \sqrt {{{\left( {80} \right)}^2} + 2 \times 9.8 \times 10} \\\\{v_2} = \sqrt {6596} \\\\{v_2} = 81.2157\,\,\,{\rm{m/s}}\end{array}\]
ANSWER
Final Velocity upon impact is \[{v_2} = 81.2157\,\,\,{\rm{m/s}}\]
We will use the conservation of energy to find the speed of ball on impact.
Step 2: Initial Energy of ball,
Kinetic energy of ball (mass $m$ and initial velocity $v_1$) before firing. \[{K_1} = \dfrac{1}{2}mv_1^2\]Potential energy of ball mass $m$ and initial height $y_1$) before firing\[{U_1} = mg{y_1}\]
Step 3: Final Energy of ball,
Kinetic energy of ball (mass $m$ and final velocity $v_2$) upon impact \[{K_2} = \dfrac{1}{2}mv_2^2\]Potential energy of ball (mass $m$ and final height $y_2$) upon impact with mass $m$ and initial height velocity $y_1$. \[{U_2} = mg{y_2}\]
Step 4: Applying conservation of Total energy
Energy before firing is equal to energy after impact\[\begin{array}{l}{K_1} + {U_1} = {K_2} + {U_2}\\\\\dfrac{1}{2}mv_1^2 + mg{y_1} = \dfrac{1}{2}mv_2^2 + mg{y_2}\end{array}\]
Step 5: Final velocity
Initial height of ball is 10 m. Upon impact height becomes 0. Intital velocity is given by 80 m/s.\[\begin{array}{l}\dfrac{1}{2}mv_1^2 + mg{y_1} = \dfrac{1}{2}mv_2^2 + mg{y_2}\\\\\dfrac{1}{2}v_1^2 + g{y_1} = \dfrac{1}{2}v_2^2 + g{y_2}\\\\\dfrac{1}{2}{\left( {80} \right)^2} + 9.8\left( {10} \right) = \dfrac{1}{2}v_2^2 + g\left( 0 \right)\\\\{v_2} = \sqrt {{{\left( {80} \right)}^2} + 2 \times 9.8 \times 10} \\\\{v_2} = \sqrt {6596} \\\\{v_2} = 81.2157\,\,\,{\rm{m/s}}\end{array}\]
ANSWER
Final Velocity upon impact is \[{v_2} = 81.2157\,\,\,{\rm{m/s}}\]