Question

总质量为24kg的气球,以2m/s的速度竖直匀速上升.当升离地面300m高处时,从气球上落下一个质量为4kg,体积

Answer 1

1.F=m1g

a=(F-m2g)/m2=(m1g-m2g)/m2=(24*9.8-20*9.8)/20=1.96m/s^2

h2=vt+0.5at^2=2*5+0.5*1.96*25=34.5m

h=h1+h2=300+34.5=334.5m

2.h3=-vt+0.5gt^2=-2*5+0.5*9.8*25=112.5m

h4=h1-h3=300-112.5=187.5m