11판/2. 직선운동

2-50 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 1. 17. 21:13
S=vt12at2, S=vt-\frac{1}{2}at^2, {S1.5=v1.5t1.512a1.5t1.52S1=v1t112a1t12 \begin{cases} S_{1.5}=v_{1.5}t_{1.5}-\frac{1}{2}a_{1.5}{t_{1.5}}^2\\ S_1=v_1t_1-\frac{1}{2}a_1{t_1}^2 \end{cases} {H1.5=(0)t1.512(g)t1.52H1=(0)t112(g)t12 \begin{cases} H_{1.5}=(0)t_{1.5}-\frac{1}{2}(-g){t_{1.5}}^2\\ H_1=(0)t_1-\frac{1}{2}(-g){t_1}^2 \end{cases} {H1.5=12gt1.52H1=12gt12 \begin{cases} H_{1.5}=\frac{1}{2}g{t_{1.5}}^2\\ H_1=\frac{1}{2}g{t_1}^2 \end{cases} H1.5H1=12gt1.5212gt12=(t1.5t1)2=1.52=94 \begin{aligned} \frac{H_{1.5}}{H_1}&=\frac{\frac{1}{2}g{t_{1.5}}^2}{\frac{1}{2}g{t_1}^2}\\ &=\(\frac{t_{1.5}}{t_1}\)^2\\ &=1.5^2\\ &=\frac{9}{4} \end{aligned} H1.5=94H1=2.25H1 \begin{aligned} H_{1.5}&=\frac{9}{4}H_1\\ &=2.25H_1\\ \end{aligned}