Question

A rocket is fired vertically upwards starting frkm rest. It accelerates at 30m/s for 4secs. At the end of 4secs it runs out of fuel but continues to rise. How long does it rise with reference to firing position?

Answer 1

t = 16.5 s

Step-by-step explanation:

First we apply first equation of motion to the accelerated motion of the rocket:

`v_(f1) = v_(i1) + at_(1)`

where,

vf₁ = final speed of rocket during accelerated motion = ?

vi₁ = initial speed of rocket during accelerated motion = 0 m/s

a = acceleration of rocket during accelerated motion = 30 m/s²

t₁ = time taken during accelerated motion = 4 s

Therefore,

`v_(f) = 0\ m/s + (30\ m/s^2)(4\ s)\\\\v_(f) = 120\ m/s`

Now, we analyze the motion rocket when engine turns off. So, the rocket is now in free fall motion. Applying 1st equation of motion:

`v_(f2) = v_(i2) + g t_(2)`

where,

vf₂ = final speed of rocket after engine is off = 0 m/s

vi₂ = initial speed of rocket after engine is off = Vf₁ = 120 m/s

g = acceleration of rocket after engine is off = - 9.8 m/s² (negative sign for upward motion)

t₂ = time taken after engine is off = ?

Therefore,

`0\ m/s = 120\ m/s + (- 9.8\ m/s^2)(t_(2))\\\\t_(2) = (120\ m/s)/(9.8\ m/s^2)\\\\t_(2) = 12.25\ s`

So, the time taken from the firing position till the stopping position is:

`t = t_(1) + t_(2)\\\\t = 4 s + 12.5 s`

t = 16.5 s