Question

A model rocket is launched with an acceleration of 22.5 m/s2 at an angle of 72.0o (measured from the horizontal). If the model rocket has a mass of 3.55 kg what would be the net horizontal and vertical forces involved

Answer 1

Let the acceleration magnitude be $a = 22.5m/s^2$ and the mass $m = 3.55kg$. Then, according to the Newton's second law, the net force acting on the model is colinear with the acceleration and has the magnitude (see the picture):

$$F = ma = 3.55\times22.5 = 79.875\space N$$

The horizontal component of this force is:

$$F_x = F\cos\theta = 79.875\times \cos72\degree \approx 24.68\space N$$

The vertical component:

$$F_y = F\sin\theta = 79.875\times \sin72\degree \approx 75.97\space N$$

Answer. Horizontal 24.68 N, vertical 75.97 N.