Question

Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial speed of 18 m/s at an angle 40 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground.
What is the vertical component of the ball’s velocity right before Sarah catches it?

Answer 1

We know that \(\Large v_y=v_{0y}t - \frac{1}{2} g t^2 = (v_0sin40)t-\frac{1}{2}gt^2\)