Physics Question about parabola projectile angle

Question

A projectile is launched over level ground at $35 \frac{m}{s}$ at an angle of $24^{\circ}$ above the horizontal. Friction is negligible. What is the velocity (magnitude and direction) of this projectile $2.5$ s after launch?

For the magnitude do they mean the overall velocity at $2.5$ seconds $(\cos(24)35)$ in or do they mean the magnitude of the displacement: $\cos(24)35 \cdot 2.5$ ?

For the direction part do you assemble a triangle of the the combined velocity and the x and y components. Then use the cos law to solve for cos of the angle? Is this completely wrong? The answer on my homework differs from mine, but maybe thats just cause my math skills are questionable. Can anyone help me out with this and solve it step for step with an answer at the end? Thanks soooooo much :)!!!!!!

Answer 1

For the magnitude of the velocity, it is $|v|=\sqrt{v_x^2+v_y^2}$, where $v_x$ and $v_y$ are the velocities in the $x$ and $y$ directions. For the direction part you should describe it as degrees above or below the horizontal. It will be $\arctan \frac{v_y}{v_x}$. As Rober Wiberg comments, you need to account for gravity changing the original velocity.

Answer 2

The key idea here is to separate the question into x and y components. So the overall velocity at $2.5$ seconds is not $\cos(24) \cdot 35$ - this is the initial x component velocity (although this does not change). The y component does change, due to the force of gravity. Use Newton to find that.

I see now that Ross answered with respect to the process of finding the magnitude and direction of the velocity. So I won't comment on that other than to emphasize the need to think component-wise.