作业帮一道大学英文数学题求解
来源：学生作业帮 编辑：拍题作业网作业帮 分类：综合作业 时间：2024/04/30 18:31:26
一道大学英文数学题求解
是一道MODELLING AND DYNAMICS的题
An inclined plane passes through the origin and makes an angle β>0 with the horizontal.A ball is thrown from the origin up the plane with speed V at an angle α>β to the horizontal.
i)Find x(t),the path of the ball,assuming that the only force acting on the ball is its weight,vertically downwards.
ii)Show that the ball lands on the plane at time t=2Vsin(α-β)/(gcosβ).
iii)Find the range r of the ball(i.e.the distance from the origin to the point at which the ball lands on the plane).
iv)Show that the maximum range is achieved when α bisects the angle between the plane and the vertical,and that the maximum range is V平方/[g(1+sinβ)].
是一道MODELLING AND DYNAMICS的题
An inclined plane passes through the origin and makes an angle β>0 with the horizontal.A ball is thrown from the origin up the plane with speed V at an angle α>β to the horizontal.
i)Find x(t),the path of the ball,assuming that the only force acting on the ball is its weight,vertically downwards.
ii)Show that the ball lands on the plane at time t=2Vsin(α-β)/(gcosβ).
iii)Find the range r of the ball(i.e.the distance from the origin to the point at which the ball lands on the plane).
iv)Show that the maximum range is achieved when α bisects the angle between the plane and the vertical,and that the maximum range is V平方/[g(1+sinβ)].
(1)
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.
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.