Question

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.

Answer 1

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,

• $v_f$ = 375 m/s

Height,

• h = 1.25 km

           = 1250 m

As we know,

→ $vf^2 = u^2+2ah$

or,

→    $a = \frac{v_f^2-u^2}{2h}$

By putting the values, we get

→       $=\frac{(375)^2-(0)^2}{2\times 1250}$

→       $=\frac{140625}{2500}$

→       $= 56.2 \ m/s^2$

Thus the above solution is correct.

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