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 negligable, 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.

the answer is suppose to be in two graphs, one for horizontal velocity and the other for vertical velocity. time along x-axis in .5sec increments.

vₓ = 10 m / s,    = 10 - 10 t

Explanation:

This is a projectile launching exercise, where the x and y axes are treated independently.

On the x axis. towards the north there is no acceleration, so the speed is constant

          vₓ = v₀ₓ

On the vertical y-axis, there is the acceleration of gravity with value (-g) is directed down, the equation is

          =   - gt

Let's use trigonometry to find the initial velocity component

          sin 45 =  / v₀

          cos 45 = v₀ₓ / v₀

           = v₀ sin45

          v₀ₓ = v₀ cos45

To find the initial velocity let's use the scope equation

         R = v₀² sin 2θ / g

         v₀ = √ Rg / sin 2θ

         v₀ = √ (20 10 / sin (2 45))

         v₀ = 14.14 m / s

With this value we find the initial velocity component

           = 14.14 sin 45

            = 10 m / s

        v₀ₓ = 14.14 cos 45

        v₀ₓ = 10 m / s

We get the equations to graph

X axis

       vₓ = 10 m / s

In this graph it is a horizontal line

Y Axis

       = 10 - 10 t

For this graph let's build a table

t (s)        (m / s)

0           10

0.5         5

1.0          0

1.5         -5

2          -10

see attached for both curves