Question

A rocket is fired vertically from the ground. It moves upward with a constant acceleration 10 m/s2 for 30s after which the fuel is consumed. After what time from the instant of firing the rocket will attain the maximum height?

Answer 1

Answer:

Explanation:

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Answer 2

Answer:

At 60 sec from the instant of firing the rocket will attain the maximum height.

Explanation:

The rocket fired vertically moves up with constant acceleration and attains the maximum height in the time of its ascent. The time of ascent is calculated using equation of motion.

As we know that, from the second equation of motion and third equation of motion, so, from here, we have –

$\bold{\begin{array}{l}{S=u t+1 / 2 a t^{2}} \\ {v^{2}=u^{2}+2 a s}\end{array}}}$

Given is that initial velocity u = 0.

Final velocity v = ?

Acceleration = 10

Time = 30 sec

$\begin{array}{l}{S=u t+1 / 2 a t^{2}} \\ {S=0 \times t+1 / 210 \times 30 \times 30}\end{array}$

S = 4500 m

$\begin{array}{l}{\underline{V}^{2}=u^{2}+2 a s} \\ {\underline{v}^{2}=0 \times 0+2 \times 10 \times 4500} \\ {\underline{v}^{2}=20 \times 4500}\end{array}$

v  = 300 m/ s

Now from the first equation of motion,  

v = u + at

300 = 0 + a x t

300 / 10 = t

t = 30 sec

The total time for ascent = time before fuel consumed and time after fuel consumed = 30 + 30 = 60 sec.