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Travel distance when an object on a slope is thrown at a certain angle

작성자 Uploader : narsha 작성일 Upload Date: 2018-09-11변경일 Update Date: 2018-09-11조회수 View : 376

Type1 Type2
Assume a condition shown in the figure below.
Letting,
initial velocity of object : V
angle of throwing the object with respect
to the horizon : θ1(counterclockwise direction is +.)
angle of slope with respect to the horizon : θ2(clockwise direction is +.)

x direction velocity = Vx = Vcos(θ1)
y direction velocity= Vy = Vsin(θ1)-gt
x direction travel distance = Lx = Vcos(θ1)t
y direction travel distance = Ly = Vsin(θ1)t - (1/2)gt^2

t = Lx/(Vcos(θ1))

Ly = -tan(θ2)Lx = Vsin(θ1)t - (1/2)gt^2
-tan(θ2)Lx = Vsin(θ1)Lx/(Vcos(θ1)) - (1/2)g(Lx/(Vcos(θ1)))^2

Lx = 2/g*(tan(θ1)+tan(θ2))*(V*cos(θ1))^(2)

Travel distance along the slope : D = (Lx^2 + Ly^2)^(1/2)
D = Lx*(1+(tan(θ2))^2)^(1/2)
D = 2/g*(tan(θ1)+tan(θ2))*(V*cos(θ1))^(2)
    *(1+(tan(θ2))^2)^(1/2) 		No Image


▶ Input data : V (m/s), θ1(deg), θ2(deg), g(m/s^2)




■ x direction travel distance : Lx




■ y direction travel distance : Ly




■travel distance along the slope : D


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※ Input data on red underlines.

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