A particle is projected from ground level with speed U at angle α to the horizontal towards a wall a distance X away. It hits the wall horizontally.
a) Find X in terms of U, α and g.
The particles speed is halved when it rebounds horizontally from the wall. It lands on the ground 2m in front of the point of projection.
b) Show that U=2√(g/ sinα conα)
Maths~M2
2007-10-15 9:14 am
回答 (1)
2007-10-15 9:47 am
✔ 最佳答案
a.)Ux(t) = UcosA, Uy(t) = UsinA-gt
where Ux and Uy are horizontal and vertical speed respectively.
As it hits the wall horizontally, Uy(T) = UsinA-gT=0, T is the hitting moment.
T=UsinA/g
X=Ux(t)T
=U^2sinAcosA/g ##
b.)
After hitting, the particle behaves as projected from hitting point with initial horizontal speed UcosA/2. [Case 1]
If the speed is not halved, i.e. speed= UcosA, then the projection will make the particle hit X (the point of projection) away. (Reversible) [Case 2]
By Case 2, Time spent = X/UcosA = UsinA/g
From Case 1, Horizontal speed = Horizontal distance/time
UcosA/2 = (X-2)/(UsinA/g)
U^2 sinAcosA/2g = X-2
X/2 = X-2
X = 4
U^2 sinAcosA/g = 4
U=2√(g/ sinA cosA) ##
參考: me
收錄日期: 2021-04-23 20:04:58
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