Question

 氣球載一重物以0.8 m/s ² 的加速度從地面升起,10 s末重物與氣球脫離,此后重物還能上升的時(shí)間為________s,重物落地時(shí)速度的大小為________m/s.(g取10 m/s ² )

Answer 1

 0.8    29.4

解析：據(jù)v=at,求得氣球上升10 s時(shí)速度v ₁ =at=0.8×10 m/s=8 m/s

設(shè)向上為正,設(shè)重物脫落后上升時(shí)間為t′,則有v ₁ =8 m/s,a=-10 m/s ² ,v _t =0

據(jù)v _t =v ₀ +at得t′= s=0.8 s

據(jù)s=v ₀ t+ at ² 解得加速上升高度為h ₁ = at ₁ ² = ×0.8×10 ² m=40 m

減速上升高度為:h ₂ =- gt′ ² =- ×(-10)×(0.8) ² m=3.2 m

則自由落體高度h=h ₁ +h ₂ =(40+3.2) m=43.2 m ，據(jù)v ² =2gh得v= = m/s=29.4 m/s.