Physics problem...... (two dimensional motion) (10 pts)

let x be the hori. disp
let y be the vert. disp
let t be the time of flight

y/x = tan 20 -------------------1 (as the strike pt located on the ramp)
x=15t cos 53 ------------------2 (h. vel = 15 cos 53, h accel = 0)
y=15t sin 53 -1/2 gt^2 -----3 (v. vel = 15 sin 53, v accel = -g g: gravity)

2 and 3 come from formula s=ut+1/2at^2

solve 3 equation for 3 unknowns

you can get the ans.

a projectile is fired with an initial velocity of 15ms-1 at 53o above the horizontal from the foot of a ramp inclined 20 o above horizontal.
設抛射體的初速為 vo(15ms-1)，抛射角為θ(53o)，則
vox = vo cosθ
voy = vo sinθ
水平方向為等速運動
x = voxt
t = x / vox ___(1)
垂直方向為沿垂上抛運動
y = voyt – 0.5gt2
(1)式代入上式
y = voy(x/vox) – 0.5g(x/vox)2 ___(2)
上式便是抛射體的軌跡。若它向傾角α(20 o)斜面上
抛射面的方程可以寫為
y = x tanα ___(3)
若物體抛在斜面上，則物體的軌跡方式(2)，和斜面的方程式(3)，兩方程式的交點便是物在的落點。

vox = vo cosθ = 15cos53 = 9..3
voy = vo sinθ = 15sin53 = 12.0

y = voy(x/vox) – 0.5g(x/vox)2
y = 12(x/9.03) – 0.5(10)(x/9.03)2
y = 1.33x - 0.0613x2 ___(4)

斜面的方程
y = x tanα
y = x tan 20
y = 0.364x ____(5)

用(4)及(5)式解
0.364x = 1.33x - 0.0613x2
0.364 = 1.33 - 0.0613x
x = 15.8m
y = x tan20
y = 7.54m

則點和原來抛射的地點相距
√(15.82 + 7.542)
= 17.5m