Question

关于篮球抛物线公式的解（超急）

Answer 1

更新1:

能不能请你转成中文- -？

Consider motion in the horizontal direction
horizontal velocity = (vo)cos(40) horizontal distance travelled by the ball = 10 m (given on the diagram) Since hroizontal distance travelled = horizontal velocity x time taken t hence
10 = (vo)cos(40) x t i.e. t = 10/[(vo).cos(40)] ----------------- (1) Consider motion in the vertical direction
vertical displacement of the ball = height of basket above ground - height of projection of the ball i.e. vertical displacement △y = (3.05 - 2) m = 1.05 m vertical velocity of ball on projection = (vo).sin(40) use equation of motion: s = ut + (1/2)at^2 with s = △y = 1.05 m
u = (vo).sin(40)
a = -g
hence
1.05 = [(vo)sin(40)]t + (1/2)(-g)t^2 Because we are referring to the same ball
using t from (1)
1.05 = [(vo)sin(40)].10/[(vo)cos(40)] - (g/2)[10/(vo)cos(40)]^2 i.e. 1.05 = 10.tan(40) - (g/2)[10/(vo).cos(40)]^2 [10/[(vo)cos(40)]^2 = (2/g) x (10.tan(40) - 1.05) 10/[(vo).cos(40)] = square-root(1.497) vo = [10/cos(40)] /1.223 m/s = 10.7 m/s