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
orth\ 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 graph
vertical velocity graph

Explanation:

Step1: Find horizontal initial velocity

The horizontal displacement \( x = v_{0x}t \), where \( x = 20\space m \), \( t = 2\space s \). So \( v_{0x}=\frac{x}{t}=\frac{20}{2}=10\space m/s \). Since there's no air resistance, horizontal velocity \( v_x = v_{0x}=10\space m/s \) (constant).

Step2: Find vertical initial velocity

The angle is \( 45^\circ \), so \( \tan(45^\circ)=\frac{v_{0y}}{v_{0x}} \), so \( v_{0y}=v_{0x}=10\space m/s \). The vertical velocity as a function of time is \( v_y = v_{0y}-gt \), with \( g = 10\space m/s^2 \), so \( v_y=10 - 10t \). At \( t = 0 \), \( v_y = 10\space m/s \); at \( t = 1\space s \), \( v_y = 0\space m/s \); at \( t = 2\space s \), \( v_y=-10\space m/s \).

For horizontal velocity plot: It's a horizontal line at \( v_x = 10\space m/s \) from \( t = 0 \) to \( t = 2\space s \).

For vertical velocity plot: It's a straight line with slope \( - 10\space m/s^2 \), starting at \( (0, 10) \), passing through \( (1, 0) \), and ending at \( (2, - 10) \).

Answer:

Horizontal velocity: constant \( 10\space m/s \) (plot as horizontal line \( v = 10 \) for \( 0\leq t\leq2 \)).
Vertical velocity: \( v_y = 10 - 10t \) (plot as line from \( (0, 10) \) to \( (2, - 10) \)).