Projectile motion
Projectile motion involves a movement with two separate components:

Vertical movement due to gravitation. Care should be taken to clearly define the positive vertical direction (upward can be positive or negative, depending on what is most convenient for the exercise).
If the orientations of the velocity and gravitation are opposite: $$\vecphy{a}=\vecphy{g}=9.81\text{ m/s}^2\approx 10 \text{m/s}^2$$. Otherwise, $$\vecphy{a}=\vecphy{g} \approx 10 \text{m/s}^2$$.
 Initial velocity due to the projection, which is usually at an angle. Assuming no air resistance, the horizontal velocity is not subject to any acceleration and is constant.
These problems can be solved by considering the velocity in the horizontal and vertical planes separately.
For projectile motion where the object is shot horizontally, the horizontal velocity $$\vecphy{v}_x$$ reduces to $$\vecphy{v}_{0x}$$, where $$\vecphy{v}_{0x}$$ is the starting velocity.
Assuming no initial vertical velocity $$\vecphy{v}_{0y}=0 \text{ m/s}$$, the vertical velocity reduces to $$\vecphy{v}_y=\vecphy{v}_{0y}+gt=gt.$$
If a projectile is shot at an angle of $$\alpha$$, the initial velocity $$\vecphy{v}_{0}=u$$ can be split into a horizontal $$(\vecphy{v}_{0x})$$ and a vertical $$(\vecphy{v}_{0y})$$ component:$$$\vecphy{v}_{0x}= \vecphy{v}_{0}\cos \alpha \quad \quad \vecphy{v}_{0y}= \vecphy{v}_{0}\sin\alpha$$$
The horizontal acceleration is zero while the vertical acceleration is given by gravitation:$$$\vecphy{a}_x=0 \quad \quad \vecphy{a}_y=\vecphy{g}\approx 10 \text{ m/s}^2$$$
The horizontal velocity component remains constant while the vertical changes due to gravitation: $$$\vecphy{v}_x= \vecphy{v}_{0}\cos \alpha \quad \quad \vecphy{v}_y= \vecphy{v}_{0}\sin\alpha\vecphy{g}t$$$
Horizontal and vertical displacement are:$$$\vecphy{s}_x=\vecphy{v}_{0}t\cos\alpha \quad \quad \vecphy{s}_y= \vecphy{v}_{0}t\sin \alpha\frac{1}{2}\vecphy{g}t^2$$$
If the initial movement is upward, then $$g$$ needs to be set to be negative.