11판/4. 2차원 운동과 3차원 운동

4-33 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 2. 6. 18:57
{A:BallB:FloorC:Elevator \begin{cases} A:\text{Ball}\\ B:\text{Floor}\\ C:\text{Elevator}\\ \end{cases} {v0AB=v0=8.00[m/s]vC=2.00[m/s]g=9.80665[m/s2] \begin{cases} v_{0A\larr B}&=v_0=8.00\ut{m/s}\\ v_{C}&=2.00\ut{m/s}\\ g&=9.80665\ut{m/s^2}\\ \end{cases} (a)\ab{a} 2aS=v2v02,2(g)(H)=(0)282 \begin{aligned} 2aS&=v^2-{v_0}^2,\\ 2(-g)(H)&=(0)^2-{8}^2\\ \end{aligned} H=32g3.263091881529371[m]3.26[m] \begin{aligned} H&=\frac{32}{g}\\ &\approx 3.263091881529371\ut{m}\\ &\approx 3.26\ut{m}\\ \end{aligned} (b)\ab{b} 2aS=v2v02,2(g)(HB)=(0)262 \begin{aligned} 2aS&=v^2-{v_0}^2,\\ 2(-g)(H_B)&=(0)^2-{6}^2\\ \end{aligned} H=18g1.835489183360271[m]1.84[m] \begin{aligned} H&=\frac{18}{g}\\ &\approx 1.835489183360271\ut{m}\\ &\approx 1.84\ut{m}\\ \end{aligned} (c)\ab{c} aAB=aAaB=g0=9.80665[m/s2] \begin{aligned} a_{A\larr B}&=a_A-a_B=-g-0\\ &=-9.80665\ut{m/s^2} \end{aligned} (d)\ab{d} aAC=aAaC=g0=9.80665[m/s2] \begin{aligned} a_{A\larr C}&=a_A-a_C=-g-0\\ &=-9.80665\ut{m/s^2} \end{aligned}