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

Answer:

t = 16.5 s

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,

$0\ m/s = 120\ m/s + (- 9.8\ m/s^2)(t_{2})\\\\t_{2} = \frac{120\ m/s}{9.8\ m/s^2}\\\\t_{2} = 12.25\ s$

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