A glaucous-winged gull,ascending straight upward at 5.20m/s,drops a shell when it is 12.5m above the ground.a) What is the magnitude and direction of the shell's acceleration just after it is released?b) Find the maximum height above the ground reach

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A glaucous-winged gull,ascending straight upward at 5.20m/s,drops a shell when it is 12.5m above the ground.a) What is the magnitude and direction of the shell's acceleration just after it is released?b) Find the maximum height above the ground reach

A glaucous-winged gull,ascending straight upward at 5.20m/s,drops a shell when it is 12.5m above the ground.a) What is the magnitude and direction of the shell's acceleration just after it is released?b) Find the maximum height above the ground reach
A glaucous-winged gull,ascending straight upward at 5.20m/s,drops a shell when it is 12.5m above the ground.a) What is the magnitude and direction of the shell's acceleration just after it is released?b) Find the maximum height above the ground reached by the shell.c) How long does it take for the shell to reach the ground?d) What is the speed of the shell at this time?
a) 9.81m/s^2
b) 13.9m
C) 2.21s
d) 16.5m/s

A glaucous-winged gull,ascending straight upward at 5.20m/s,drops a shell when it is 12.5m above the ground.a) What is the magnitude and direction of the shell's acceleration just after it is released?b) Find the maximum height above the ground reach
一只鸟以5.2m/s的速度垂直向上飞,在12.5m的高度丢了一个贝壳下来
a.贝壳刚被丢下来的时候的加速度大小和方向
b.贝壳到达的最高高度
c.贝壳用多久落地
d.贝壳落地时的速度
a.就是重力加速度.9.81m/s^2,垂直向下
b.贝壳被丢下来之后那一刻速度还是5.2m/s...但是因为重力加速度,速度会不断减慢,直到0,这期间用时5.2/9.81=0.53s,这段距离是5.2*0.53/2=1.38m
总高度就是12.5+1.38=13.9m
c.这一题,我做出来答案不一样.
距离=9.81*（t^2)/2
t=sqrt(13.9*2/9.81)=1.68s
d.速度从最高处0开始,落地速度=重力加速度*时间=9.81*1.68=16.5m/s

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