Question

如题所示；A、B两球之间用长为6m的绳子相连,将两球相隔0.8s先后从同一高度以4.5m/s的水平速度抛出,则经过多长时

Answer 1

由平抛运动规律可得 ：xA=v0tm yA=1/2×gt^2

xB =v0(t-0.8),yB=1/2×g(t-0.8)^2

l^2 =(xA-xB)^2 +(yA-yB)^2

代入数字得

36 =(4.5×0.8)^2 +[1/2×10×(1.6t-0.64)]^2

解得 t=1s,

xA=v0t=4.5m

yA=1/2×gt^2=5m

sA^2 = xA^2 +yA^2 = 4.5^2 +5^2 = 45.25

sA =6.73m