Question

A rocket of 3 × 10⁶ Kg mass takes off from the launching pad and acquires a vertical velocity of 1 Km S-1 at an altitude of 25Km. Calculate (a) the potential energy, and (b) the kinetic energy. (Take g=10m s-1 ).​

Answer 1

Answer:

potential energy

• $formula \: = \: mgh$
• m = 3 × 10⁶ kg
• g = 10 m/ s²1
• h = 25 km

Therefore potential energy

$\: 3 \times {10}^{6} \: \times 10 \: \times 25$

$potential \: energy \: = 75 \: \times {10}^{7}$

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Now to calculate Kinetic energy

$formula \: of \: kinetic \: energy \\ \\ = 1 \div 2(m \times {v)}^{2}$

• m = 5 × 10⁷kg
• v = 1 km / s
• (first convert 1 km / s into km/h)
• therefore v = 3600 km / h

$= \: 3 \times {10}^{6} \times {3600}^{2} \div 2$

therefore

$kinetic \: energy \: = \: 5 \times {10}^{6} \times 6480000$

hence

$kinetic \: energy \\ \\ = 324 \times {10}^{11}$

here is your answer and it's verification

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