contestada

A student launches a small rocket which starts from rest at ground level. At a height of h = 1.25 km the rocket reaches a speed of vf = 375 m/s. At that height the rocket runs out of fuel, so there is no longer any thrust propelling it. Take the positive direction to be upward in this problem.

Respuesta :

Cricetus

The acceleration of the rocket will be "56.2 m/s²".

According to the question,

The initial speed during launch,

  • u = 0 m/s

The speed at fuel running out point,

  • [tex]v_f[/tex] = 375 m/s

Height,

  • h = 1.25 km

           = 1250 m

As we know,

→ [tex]vf^2 = u^2+2ah[/tex]

or,

→    [tex]a = \frac{v_f^2-u^2}{2h}[/tex]

By putting the values, we get

→       [tex]=\frac{(375)^2-(0)^2}{2\times 1250}[/tex]

→       [tex]=\frac{140625}{2500}[/tex]

→       [tex]= 56.2 \ m/s^2[/tex]

Thus the above solution is correct.

Learn more:

https://brainly.com/question/11038449

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