A rocket moves straight upward with zero initial velocity and with an acceleration 20\ {m/s}². It runs out of fuel and stops accelerating at the end of 5^{{th }}\ {sec}. It reaches a maximum height and falls back to the earth. The speed when it hits the ground is ( Take g=10\ {m/s}²)

Correct Answer is : 506 m/s50{6}\ {m/s}506 m/s

Correct Answer is : 506 m/s50{6}\ {m/s}506 m/s

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TS EAMCET 19-July-2022 Shift 2 Solved Paper

Section: Physics

Question : 83 of 160

Marks: +1, -0

20\ \mathrm{m/s}^2

5^{\text{th }}\ \mathrm{sec}

(

g=10\ \mathrm{m/s}^2)

Solution:

Initial velocity of rocket,

u=0\ \mathrm{m/s}

acceleration,

a=20\ \mathrm{m/s}^2

Upward velocity of rocket at the end

5\ \mathrm{s}

v\;=u+at

\;=0+20 \times 5=100\ \mathrm{m/s}

Distance travelled by rocket in

5\ \mathrm{s}

s_1=\;\frac{1}{2} a t^2=\;\frac{1}{2} \times 20 \times 5^2=250\ \mathrm{m}

After

5\ \mathrm{s}

, rocket moves under the effect of gravitational acceleration and with initial velocity of

100\ \mathrm{m/s}

\therefore

Distance travelled by rocket in this situation is

s_2

, then using equation,

v^2=u^2-2 g s_2

\Rightarrow 0=100^2-2 g s_2 \Rightarrow s_2=\;\frac{100^2}{2 g}=\;\frac{100 \times 100}{2 \times 10}=500\ \mathrm{m}

\therefore

Total upward distance covered by rocket,

s=s_1+s_2=250+500=750\ \mathrm{m}

Now, rocket starts falling from this height and if

v'

be the velocity of rocket when it reaches at ground, then

(v')^2\;=u^2+2 g s

\;=0+2 \times 10 \times 750=15000

\Rightarrow \;\; v'\;=\sqrt{15000}=50\sqrt{6}\ \mathrm{m/s}

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