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 horizont
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![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 horizont](/uploads/image/z/5970513-57-3.jpg?t=%E4%B8%80%E9%81%93%E5%A4%A7%E5%AD%A6%E8%8B%B1%E6%96%87%E6%95%B0%E5%AD%A6%E9%A2%98%E6%B1%82%E8%A7%A3%E6%98%AF%E4%B8%80%E9%81%93MODELLING+AND+DYNAMICS%E7%9A%84%E9%A2%98An+inclined+plane+passes+through+the+origin+and+makes+an+angle+%CE%B2%3E0+with+the+horizontal.A+ball+is+thrown+from+the+origin+up+the+plane+with+speed+V+at+an+angle+%CE%B1%3E%CE%B2+to+the+horizont)
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 horizont
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)].
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 horizont
(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.
分太低。