已知:Vt = 1/4Vo,h = 3m
求:1.Vo ; 2.t
根據(jù)公式
Vo^2-Vt^2 = 2gh
Vo^2-1/16Vo^2 = 2*10*3
15/16Vo^2 = 60
Vo^2 = 60*16/15
Vo = 8 m/s (Vt = 2m/s)
2.
如果問的是物體從拋出到回到手里經(jīng)過多長(zhǎng)時(shí)間,則 H = 0
H = Vo*t-1/2g*t^2 = 0
5t^2= 8t
t = 8/5s
這是物體拋出后8/5秒回到手中.
如果問的是物體到達(dá)B點(diǎn)后再經(jīng)過多長(zhǎng)時(shí)間物體落回A處則 H = -3m
H = Vt*t - 1/2g*t^2 = -3
5t^2-2t-3 = 0
(t-1)(5t+3) = 0
t1 = 1s
t2 = -3/5s 無意義舍去
故再經(jīng)過1秒時(shí)間物體落回A處.