5. A rocket of massv3x10⁶ kg takes off from a launching pad and acquires a vertical velocity of 1km/s at an altitude of 25 km. calculate (a) the potential energy and (b) kinetic energy. (g = 9.8m/s²)

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Solution

$i) P . E = m g h = 3 \times 10^{6} \times 9.8 \times 25 \times 10^{3} = 735 \times 10^{9} J i i) K . E = \frac{1}{2} m v^{2} = \frac{1}{2} \times 3 \times 10^{6} \times {(10^{3})}^{2} = 1.5 \times 10^{12} J$

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Similar questions

Q. A rocket of mass

3 \times 10^{6} k g

takes off from a launching pad and acquires a vertical velocity of

1 k m s^{- 1}

and an altitude of

25 k m

. Calculate its (a) potential energy (b) kinetic energy.

Q. A rocket of mass

3 \times 10^{6}

kg takes off from a launching pad and acquires a vertical velocity of 1 km/s at an altitude of 25 km. The kinetic energy is

Q. A rocket of mass

3 \times 10^{6} k g

takes off from a launching pad and acquires a vertical velocity of 1 km/s at an altitude of 25 km. The kinetic energy of the rocket is: (Assume g is constant and having a value

10 m / s^{2}

Q. A rocket, set for vertical firing, weighs

50 kg

and contains

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of fuel. It can have maximum exhaust velocity of

2 km/s

. What should be its minimum rate of fuel consumption, to just lift it off the launching pad ?

Take

g = 9.8 {m/s}^{2}

The diagram given below shows a ski jump. A skier weighing 60 kgf stands at A at the top of the ski jump. He moves from A and takes off for his jump at B.

(a) Calculate the change in the gravitational potential energy of the skier between A and B.

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5. A rocket of massv3x106 kg takes off from a launching pad and acquires a vertical velocity of 1km/s at an altitude of 25 km. calculate (a) the potential energy and (b) kinetic energy. (g = 9.8m/s2)

i) P.E = mgh =3×106×9.8×25×103=735 ×109 Jii) K.E =12mv2=12×3×106×1032=1.5×1012 J

5. A rocket of massv3x10⁶ kg takes off from a launching pad and acquires a vertical velocity of 1km/s at an altitude of 25 km. calculate (a) the potential energy and (b) kinetic energy. (g = 9.8m/s²)

$i) P . E = m g h = 3 \times 10^{6} \times 9.8 \times 25 \times 10^{3} = 735 \times 10^{9} J i i) K . E = \frac{1}{2} m v^{2} = \frac{1}{2} \times 3 \times 10^{6} \times {(10^{3})}^{2} = 1.5 \times 10^{12} J$