A projectile is launched at some angle...

A projectile is launched at some angle to the horizontal with some initial speed and air resistance is negligible.

(a) What is its acceleration in the vertical direction? (Let up be the positive direction.)
(b) What is its acceleration in the horizontal direction?

Units for both should be in m/s^2

I know that the projectile is a freely falling body, but I have no clue how to solve this question. Thank you!

providing that it is on the earth, vertical acceleration is always 9.81m/s^2, and because there is no air resistance, horizontal acceleration is 0m/s^2

there is horizontal no acceleration, because no 'thing' is getting in the way to exert a force to cause the acceleration, because when a physicist says 'negligible air resistance' what they really mean is 'no air resistance'.

hope this helps :-)

After the projectile is launched, the only force acting on it is gravity, since the air resistance is assumed to be negligible. Gravity acts downward, with a magnitude of 10 m/s^2 (more or less, assuming we're near the surface of the Earth, and that we don't travel far enough that the curvature of the Earth comes into play).

So, since we defined up as the positive vertical direction, the acceleration in the vertical direction is -10 m/s^2.

There is no force acting in the horizontal direction, so there is no acceleration in the horizontal direction.

Does that make sense?

The projectile is subject to gravity, therefore the acceleration in the vertical direction is g. On Earth, this is 9.81m/s^2. Because it is pulled towards Earth (down), this is -9.81m/s^2.

Once it is launched, there are no forces acting in the horizontal plane, therefore there is no acceleration in the horizontal direction. Therefore the acceleration is 0m/s^2.