
Ball shot into the air
A ball is shot into the air. At a height of 9.1m, its velocity is v = (7.6i + 6.1j) m/s, with i horizontal and j upward. To what maximum height does the ball rise? What total horizontal distance does the ball travel? What are the magnitude and angel (below the horizontal) of the balls velocity just before it hits the ground?
Could somebody please explain to me how to do this and the thought process behind it?

for max height, use $\displaystyle \Delta y = \frac{v_{0y}^2}{2g}$
for horizontal displacement, you'll need $\displaystyle \Delta x = v_x t$ and
$\displaystyle \Delta y = v_{0y}t  \frac{1}{2}gt^2$. remember that $\displaystyle v_x$ remains constant.
for the angle of impact, $\displaystyle \theta = \arctan\left(\frac{v_{fy}}{v_x}\right)$ and $\displaystyle v_{fy} = v_{0y}  gt$ or $\displaystyle v_{fy} = \sqrt{v_{0y}^2  2 g \Delta y}$