大学数学题求解！！(英文题目)
发布网友
发布时间：2022-04-22 13:56
共2个回答
热心网友
时间：2023-07-10 00:49
using the coordinate system as described,
the path of the ball is determined by
y=Vsinα·t-1/2·g·t^2
x=Vcosα·t
(2)
rotate the original coordinate system by an angle β to establish a new one, which preserves the orgin but uses the inclined plane as the X axis
based on results from question 1, we have the new path of the ball as:
y'=Vsin(α-β)·t-1/2·gcosβ·t^2
x'=Vcos(α-β)·t-1/2·gsinβ·t^2
when the ball lands on the plane, y'=0
hence
t=2Vsin(α-β)/(gcosβ)
(3)
further to the results of question 2, when the ball lands on the inclined plane,
x'=Vcos(α-β)·2Vsin(α-β)/(gcosβ)-1/2·gsinβ·[2Vsin(α-β)/(gcosβ)]^2=2·V^2·[sin(α-β)cos(α-β)-sin(α-β)·sin(α-β)tgβ]/(gcosβ)
x' reaches its maximum when its derivative equals to zero
ie. derivative of sin(α-β)cos(α-β)-sin(α-β)·sin(α-β)tgβ equals to zero.
Hence cos[2(α-β)]-sin[2(α-β)]tgβ=0
ctg[2(α-β)]=tgβ
α-β=(90-β)/2
That is to say, maximum range is achieved when α bisects the angle between the plane and the vertical
By plugging α-β=(90-β)/2 into the x' formulation,
you get the maximum.
热心网友
时间：2023-07-10 00:49
