During a basketball practice, Jen throws the ball in the air with an initial velocity of 11 m/s, at an angle of 35º, and from a height of 1.4 m. At the same moment Isabelle begins running toward the ball with a constant velocity of 5 m/s, Isabelle catches the ball on its way down at a height of 2m. What is the distance between the two players when the ball is thrown?

Answer 1

the height of the ball is
h(t) = 1.4 + 11 sin35º t - 4.9t^2
it has fallen to a height of 2m at time t=1.18
Isabelle has covered 1.18 * 5 = 5.90 m

I have no idea where Isabelle was when the ball was thrown.