Question

A ball is thrown at a 30 degree angle above the horizontal with a speed of 10 ft/s. After 0.50s the horizontal component of it's velocity will be ____?

Answer 1

The horizontal component of the velocity of the ball doesn't change with time. So the horizontal component of the ball is
\[10*\cos(30)=10(\sqrt{3}/2)=5\sqrt{3}\]

Answer 2

Assuming air resistance is being ignored... it will either be 0 (if the ball has landed) or the same as its initial horizontal component. First and foremost:
\[\Large 10{\rm{ft}} = 3.05{\rm{m}}\]And then break the ball's initial velocity into its components:
\[\Large \begin{array}{l}
{v_{xi}} = {v_i}\cos \theta = 3.05 \cdot \cos 30 = 2.64{\rm{m/s}}\\
{v_{yi}} = {v_i}\sin \theta = 3.05 \cdot \sin 30 = 1.53{\rm{m/s}}\\
\end{array}\]
Now calculate how long the ball will be airborn for given its initial vertical velocity and the constant acceleration in the opposite direction of 9.81m/s/s.