The escape velocity for a projectile on the Earth's surface is 11.2 km/s. If a body is projected at twice this velocity, what happens? MCQ with Answer and Explanation

Question

The escape velocity for a projectile on the Earth's surface is 11.2 km/s. If a body is projected at twice this velocity, what happens?

A. It escapes Earth's gravity and retains a finite positive velocity in interstellar space.

B. It returns to Earth after a long time.

C. It escapes Earth's gravity but eventually comes to rest in deep space.

D. It orbits the Earth indefinitely.

Answer: Option A

Solution (By JKSSB Mock Tests)

Escape velocity (Ve) is the minimum velocity needed to overcome Earth's gravitational pull so that final velocity at infinity is zero. If launched with velocity v > Ve, the body will escape and, due to conservation of energy, retain a residual speed (v_infinity) at an infinite distance, calculated by v_infinity = √(v² - Ve²). Here, v = 2Ve, so it retains speed.

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A stone is dropped from a tower. The distance covered by it in the last second of its fall is equal to the distance covered in the first three seconds. The height of the tower is approximately: (g = 10 m/s²)

A. 80 m

B. 180 m

C. 125 m

D. 200 m

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Correct Answer: Option C

Explanation:
Distance in first 3 seconds: s₃ = ½gt² = ½×10×9 = 45 m. Let total time be n seconds. Distance in nth second: sₙ = u + g(2n-1)/2 = 0 + 5(2n-1) = 10n - 5. Given sₙ = 45 m ⇒ 10n - 5 = 45 ⇒ n = 5 s. Total height h = ½gn² = ½×10×25 = 125 m. This problem combines equation of motion with logical reasoning about time intervals. Exam tip: For free fall from rest, distance in nth second = 5(2n-1) meters when g=10 m/s². Such numerical problems test application skills in competitive exams.

This question belongs to: Science Physics