Question

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 Horizontal velocity Vertical velocity 15 15 10 10 5 5 -5 -5 -10 -10 -15 0.0 -15 0.0 0.5 1.0 1.5 2.0 0.5 1.0 1.5 2.0 t(s) t(s) vy (m/s) u

Answer 1

If a particle is projected with speed u and at an angle $\theta$ with horizontal, then time taken by it to reach ground is called time of flight T, which is given by formula:

2u sin() T =

In the question, it is given that time of flight is 2 seconds and angle is 45 degrees. So,

2= 2u sin(45) 10

2 - 2u(1/12) 10

10 =

u= 10V2 m/s

Now, if u is speed of projectile and $\theta$ is the angle of projection then,

Horizontal component of initial velocity is

uy = u cos(@

Vertical component of initial velocity is

tu, = 11 sin(0)

Here

Un = 10/2 cos(45°) = 10 m/s

ar = 0 since acceleration is in vertically downward direction

Using

V = U = 10 m/s which is a constant at any value of time.

And

$u_y = 10\sqrt{2}\sin(45^{\circ}) = 10 $ m/s$

Using

Vy = Uy + ayt

So,

$$ at t = 0.5, v_y = 10 - 0.5\times10 = 5$

$$ at t = 0.5, v_y = 10 - 0.5\times10 = 5$

at t = 1, Vy = 10 - 10 = 0

$$ at t = 1.5, v_y = 10 - (1.5\times10)= -5$

at t = 2, V, = 10 - (2 x 10) = -10

So, the corresponding graphs for Horizontal component and vertical component of velocities will be as shown:

Horizontal velocity Vertical velocity (SAU) vy(m/s) 0.5 1.5 0.5 10 1.5 1.0 t(s)