Se lanza un cohete en un ángulo de 53° sobre la horizontal con una rapidez inicial de 100 m/s. El cohete se mueve por 3.00 s a lo largo de su línea inicial de movimiento con una aceleración de 30.0 m/s2
. En este momento, sus motores fallan,
y el cohete procede a moverse como un proyectil. Determine: (a) la altitud máxima que alcanza el cohete, (b) su tiempo
total de vuelo y (c) su alcance horizontal

Explanation:

v = u + at

v₃ = 100 +30.0(3.00) = 190 m/s

s = vt + ½at²

y₃ = (100sin53)(3.00) + ½(30sin53)(3.00²) = 347.4 m

x₃ = (100cos53)(3.00) + ½(30cos53)(3.00²) = 261.8 m

a) v² = u² + uas  

s = (v² - u²) / 2a

ymax = 347.4 + (0² - (190sin53)²) / (2(-9.80)) = 1,522 m

b) t₁ = 3.00 s

   t₂ = (190sin53) / 9.80 = 15.5 s

   t₃ = √(2(1522) / 9.80) = 17.6 s

t = 3.00 + 15.5 + 17.6 = 36.1 s

c) xmax = 261.8 + (190cos53)( 15.5 + 17.6) = 4,047 m

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.

momentum of soft ball three times less than baskey ball