Emmy kicks a soccer ball up at an angle of 45° over a level field. She watches the ball's trajectory and notices that it lands, two seconds after being kicked, about 20 m away to the north. Assume that air resistance is negligible, and plot the horizontal and vertical components of the ball's velocity as a function of time. Consider only the time that the ball is in the air, after being kicked but before landing. Take "north" and "up" as the positive x ‑ and y ‑directions, respectively, and use g≈10 m/s2 for the downward acceleration due to gravity.

v_y = 10 - 10t

v_x = 10 m/s

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

it should be -8.6 cm hope this out

15.3 cos 65 degrees trig