An object is thrown with an initial velocity v0 forming an angle θ with an inclined plane, which a in turn it forms an α-angle α with a horizontal plane.
a) find the distance x from the launch point to the point where the object falls. x = f (v0, θ, α, g)
b) find the distance for v0 = 20m / s, θ = 53o , α= 36 ° , g = 9.8m / s2

Refer to the figure shown below, which is based on the given figure.

d = the horizontal distance that the projectile travels.
h = the vertical distance that the projectile travels.

Part A
From the geometry, obtain
d = X cos(α)                     (1a)
h = X sin(α)                      (1b)

The vertical and horizontal components of the launch velocity are respectively
v = v₀ sin(θ - α)               (2a)
u = v₀ cos(θ - α)             (2b)

If the time of flight is t, then
vt - 0.5gt² = -h
or
0.5gt² - vt - h = 0             (3a)
ut = d                                (3b)

Substitute (1a), (1b), (2a), (2b) (3b) into (3a) to obtain


Hence obtain

The non-triial solution for X is




Part B
v₀ = 20 m/s
θ = 53°
α = 36°

sinα + cosα tan(θ-α) = 0.8351
cosα/[v₀ cos(θ-α)] = 0.0423

X = 0.8351/(4.9*0.0423²) = 101.46 m

 X = 101.5 m

Iagree c would be the answer