NDA (I) 2022 General Ability Test · Question 69 of 150

Correct — C, 4 s. At maximum height the upward velocity becomes zero. Using v = u − gt with v = 0 gives t = u/g = 40 ÷ 10 ≈ 4 s (taking g ≈ 10 m/s²; with g = 9·8 it is about 4·08 s).

• (a)2 s — 2 s would correspond to u = gt = 10 × 2 = 20 m/s, half the given launch speed of 40 m/s.
• (b)3 s — 3 s gives u = 30 m/s, which does not match the stated 40 m/s.
• (d)5 s — 5 s implies u = 50 m/s; it is also closer to the total up-and-down flight time, not the time to reach the top.

For a body thrown straight up, gravity decelerates it at g until, at the highest point, its velocity is momentarily zero. The time to reach that point is t = u/g, where u is the launch speed. The descent takes an equal time, so the total flight time is 2u/g.

Set the final velocity to zero and solve t = u/g. Using g ≈ 10 makes the arithmetic 40/10 = 4 s. A common slip is to quote the total flight time (about 8 s) instead of the time to the top.

• At maximum height the instantaneous velocity is zero.
• Time to reach the top: t = u/g = 40/10 = 4 s.
• Maximum height reached: H = u²/(2g) = 1600/20 = 80 m.
• Time of ascent equals time of descent, so total flight time is 2u/g ≈ 8 s.

• Reporting the total flight time instead of the time to maximum height.
• Forgetting that velocity is zero (not maximum) at the top.

A plug-in of t = u/g, often paired with a follow-up for the maximum height H = u²/2g.

No directly related past PYQ was found.

• practice — not a real PYQ

  A ball thrown vertically up returns to the thrower after 8 s. Its initial speed (g = 10 m/s²) was

  • (a)20 m/s
  • (b)40 m/s
  • (c)80 m/s
  • (d)10 m/s

  Answer(b) 40 m/s — total time 2u/g = 8 s gives u = 40 m/s.

• practice — not a real PYQ

  The maximum height reached by a body thrown up at 20 m/s (g = 10 m/s²) is

  • (a)10 m
  • (b)20 m
  • (c)40 m
  • (d)5 m

  Answer(b) 20 m — H = u²/2g = 400/20 = 20 m.