在塔顶上将一物体竖直向上抛出,抛出点为A,物体上升的最大高度为20m

发布网友发布时间：2023-08-27 11:54

共1个回答

热心网友时间：2024-08-13 12:19

x = at^2/2 = 20m = 10m/s^2*t^2 / 2
得t = 2(物体上升至最高点的时间)
则v0(起抛速度) = at = 10m/s^2*2s = 20m/s

10m = v0t - at^2/2 = 20t1 - 5t^2
解得t1 = 0.58s,t2 = 3.41s

x = v0t + at^2/2 = 10m = 20t + 5t^2
t1 = 4.45s,t2 = -0.45s，t2不符合实际舍去t2

所以时间为0.58s或3.41s或4.45s

热心网友时间：2024-08-13 12:19

x = at^2/2 = 20m = 10m/s^2*t^2 / 2
得t = 2(物体上升至最高点的时间)
则v0(起抛速度) = at = 10m/s^2*2s = 20m/s

10m = v0t - at^2/2 = 20t1 - 5t^2
解得t1 = 0.58s,t2 = 3.41s

x = v0t + at^2/2 = 10m = 20t + 5t^2
t1 = 4.45s,t2 = -0.45s，t2不符合实际舍去t2

所以时间为0.58s或3.41s或4.45s

热心网友时间：2024-08-13 12:19

x = at^2/2 = 20m = 10m/s^2*t^2 / 2
得t = 2(物体上升至最高点的时间)
则v0(起抛速度) = at = 10m/s^2*2s = 20m/s

10m = v0t - at^2/2 = 20t1 - 5t^2
解得t1 = 0.58s,t2 = 3.41s

x = v0t + at^2/2 = 10m = 20t + 5t^2
t1 = 4.45s,t2 = -0.45s，t2不符合实际舍去t2

所以时间为0.58s或3.41s或4.45s

声明：本网页内容为用户发布，旨在传播知识，不代表本网认同其观点，若有侵权等问题请及时与本网联系，我们将在第一时间删除处理。E-MAIL:11247931@qq.com