Theheightofahelicopterabovethegroundisgivenbyh=2.85t3,wherehisinmetersandtisinseconds.Att=1.90s,thehelicopterreleasesasmallmailbag.Howlongafteritsreleasedoesthemailbagreachtheground?那个2.85t是立方就是

Question

The height of a helicopter above the ground is given by h = 2.85t3,where h is in meters and t is in seconds.At t = 1.90 s,the helicopter releases a small mailbag.How long after its release does the mailbag reach the ground?
那个2.85t是立方 就是2.85t的立方

Answer 1

首先,直升机的高度是一个与t的3次方成比例的变量,可见包裹的竖直方向初始速度不为零.必须求出这个速度,再结合其自由落体运动及初始竖直方向的速度、投掷时的高度解答出落地所需时间.h=2.85t³,v=-h'=-8.55t².初始时的高度h0=2.85*1.9³=19.54815；竖直方向速度v0=-30.8655.
h=v0t+1/2at²,代入得19.54815=-30.8655t+1/2*9.8*t²,解这个方程得到的解即所需要的时间.如果我没有解错的话,t=6.814935,另一个是负值,抛弃.