Question

A 0.145 kg baseball is thrown at launch angle of 30° and strikes the ground at 18 m/s. How fast would it be moving when it reaches the ground if its launch angle were 45°? Ignore air resistance.

Answer 1

The speed at which you launch the ball is the same as the speed at which it lands. Because we assume there is no friction there is no energy loss, so the initial kinetic energy has to be conserved. This also doesn't depend on the angle. Changing the angle will only change how far the ball will fly and what maximum height it will reach.
The y and x components of the velocity vector are changing with the angle.
I will assume that velocity that is given is y component.
$v_y=v_o\sin(30)\\ v_0=\frac{18}{sin(30)}\\ v_0=36\frac{m}{s}$
If we launch at the 45-degree angle y component is:
$v_y'=v_0\sin(45)\\ v_y'=25.5\frac{m}{s}$