Answer:
v_y = 10 - 10t
v_x = 10 m/s
Step-by-step explanation:
This is a projectile launch where the x axis and y axis will be treated independently.
Now, towards the north on the x-axis, there will be no acceleration and so the speed is constant
So, vₓ = v₀ₓ
Whereas, on the vertical y - axis, the acceleration due to gravity with be negative since it's in a downward direction.
Thus, the equation is;
v_y = v_oy - gt
Now, the initial velocity component will be;
cos 45 = v₀ₓ/v₀
And sin 45 = v_₀y/v₀
Thus, we have;
v₀ₓ = v₀(cos 45)
Also, v_oy = v₀(sin 45)
Now, the initial velocity would be gotten from the equation of range which is;
R = (v₀² × sin 2θ)/g
Making v₀ the subject, we have;
v₀ = √(Rg/sin 2θ)
We are given;
R = 20 m
g = 10 m/s²
θ = 45°
Thus;
v₀ = √[20 × 10/(sin (2 × 45))]
v₀ = √200
v₀ = 14.14 m/s
Thus;
v₀ₓ = 14.14(cos 45) = 10 m/s
v_oy = 14.14(sin 45) = 10 m/s
Earlier we saw that;
v_y = v_oy - gt
Thus;
v_y = 10 - 10t
Also,we saw that;
vₓ = v₀ₓ
Thus;
v_x = 10 m/s
For the graph, we will use times of t = 0, 0.2, 0.4, 0.6, 0.8, 1.0, 1.2
Thus, for each of those values of t, we will have the following values of v_oy
t (s) v_oy (m / s)
0 10
0.2 8
0.4 6
0.6 4
0.8 2
1.0 0
1.2 -2
Graph is attached