The problem is as follows:
A sphere is launched with an initial speed of $50\,\frac{m}{s^2}$ as indicated in the diagram from below. Assume that the acceleration due gravity is $g=10\,\frac{m}{s^2}$ and the angle with the horizontal is $53^{\circ}$. Given these conditions find the tangential acceleration after two seconds produced the launching of the sphere.
This problem doesn't have alternatives.
I'm confused exactly how should I approach this problem:
The only thing which I could spot is that the components would be in the form of:
$a_n=g\cos \omega$
$a_t=g\sin \omega$
where $\omega$ is the launch angle.
But I don't know exactly how to relate this with what would be happening after two seconds from launching?. I really would appreciate an answer which can show where are the vectors for the acceleration and the angles, because I dont know how to identify them properly. Can someone help me with this matter?. Please!!.
The launch angle is understood as $\tan(53^{\circ}) = \dfrac45~$, as in the $3$-$4$-$5$ Pythagorean triple.
The velocity vector as a function of time $$\vec v = (v_0 \frac35,\, v_0 \frac45 - gt) = (30,\, 40 -10t)$$
At $t = 2$ we have $\vec v = (30,\,20)$
Visually this roughly looks like the velocity where the position is the second green ball from the left, with the velocity pointing towards upper-left.
The tangential component of downward $g$ is $$g\cdot \frac{20}{\sqrt{(20)^2 + (30)^2 }} = g\cdot \frac2{\sqrt{13}} = \frac{20}{\sqrt{13}} \approx 5.547 ~\mathrm{m}/\mathrm{s}^2$$ The direction of this $5.547$ is opposite (at this particular moment) to the velocity, towards lower left.