The time taken before the two soccer players are 24.7 m apart is 3 seconds.
Velocity of Jane relative to marry is 4.4 m/s in y-direction and 6.95 m/s in x direction.
The distance between Jane and Mary after 4.08 seconds is 33.58 m.
<h3>
Resultant velocity of Mary and Jane</h3>
Vertical component of the velocity;
Jane: Vy1 = V sinθ
Vy1 = 5.07 m/s x sin(60.2)
Vy1 = 4.4 m/s
Horizontal components of the velocity;
Jane: Vx1 = V cos θ
Vx1 = 5.07 m/s x cos60.2
Vx1 = 2.52 m/s
mary: Vx2 = 4.43 m/s
total horizontal components of the velocity = 4.43 m/s + 2.52 m/s = 6.95 m/s
<h3>Resultant velocity</h3>
V = √Vy² + Vx²
V = √(4.4² + 6.95²)
V = 8.23 m/s
<h3>Time taken before they are 24.7 m apart</h3>
t = 24.7 m / 8.23 m/s
t = 3 seconds
Velocity of Jane relative to marry is 4.4 m/s in y-direction and 6.95 m/s in x direction.
<h3>Distance between Jane and Mary after 4.08 seconds</h3>
d = vt
d = 8.23 m/s x 4.08 s
d = 33.58 m
Learn more about relative velocity here: brainly.com/question/17228388
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