NDA (II) 2016 General Ability Test · Question 73 of 150

Correct — C, 2·57 s. A ball thrown straight up keeps its upward motion only until gravity has cancelled the speed it started with. At the highest point its velocity is momentarily zero, though the acceleration is still 9·8 m/s² downward throughout. Put that into v = u − gt with v set to zero and the time comes out as t = u/g. Here u is 25·2 m/s and g is 9·8 m/s², so t = 25·2 ÷ 9·8 = 2·5714 s, which rounds to 2·57 s. The arithmetic is worth noticing: 25·2 is not a random figure, it is 9·8 multiplied by 2·571, so the setter has chosen numbers that divide cleanly enough to be done in the head. Two by-products follow at once and are often asked in the same breath — the ball reaches a height of u²/2g, about 32·4 m, and returns to the ground after twice this time, 5·14 s.

• (a)5·14 s — This is the total time of flight, up and back down again, which is exactly twice the time to the top. The question asks only for the upward leg, so this answer solves a different question. The same figure also arises from dividing by g/2 instead of g.
• (b)3·57 s — No correct route produces this. It is a look-alike of the right value with the leading digit altered, offered to catch a candidate who half-remembers the answer instead of computing it.
• (d)1·29 s — This is 25·2 divided by 19·6 — the result of using 2g in place of g, or equivalently of halving the correct answer. Doubling g has no justification in this problem; the ball is decelerating at exactly 9·8 m/s².

For motion under gravity alone, the three equations of uniformly accelerated motion apply with a equal to g. Taking upward as positive, a ball thrown up with initial speed u has velocity v = u − gt, height h = ut − ½gt² and v² = u² − 2gh. At the top the velocity is zero but the acceleration is not, which is the single most misunderstood point in the topic. Air resistance is neglected in all such problems unless the question says otherwise.

Work these items by asking what is zero rather than by hunting for a formula. Here the phrase 'highest point' means the velocity has fallen to zero, and the equation containing v, u, g and t is the one to use. The distractors are constructed around the standard slips, so a quick check protects you: if your answer is double the correct one you have computed the whole flight, and if it is half you have doubled g. A useful sanity estimate is that gravity removes roughly ten metres per second of speed every second, so a ball starting at about 25 m/s must take a little over two and a half seconds to stop rising.

• At the highest point of the flight the velocity is momentarily zero while the acceleration remains 9·8 m/s² downward.
• The time to reach the top is u/g, which here is 25·2 ÷ 9·8 = 2·57 s.
• The maximum height reached is u²/2g, about 32·4 m for this throw.
• For a body thrown up and returning to the same level, the total time of flight is 2u/g, twice the time to the top.
• The upward and downward journeys are mirror images, so the ball returns to the ground with the same speed it started with.

• Giving the total time of flight when only the time to the highest point is asked for.
• Believing that acceleration is zero at the top because velocity is zero there.

NDA GAT sets one or two short numerical physics items each session, and motion under gravity is the most frequent of them, so practise until t = u/g and h = u²/2g come without thought.

• practice — not a real PYQ

  A ball thrown vertically upward returns to the thrower's hand after 6 seconds. Taking g as 10 m/s², the speed with which it was thrown is

  • (a)15 m/s
  • (b)30 m/s
  • (c)60 m/s
  • (d)20 m/s

  Answer(b) 30 m/s — the time up is half of 6 s, so u = g × 3 = 30 m/s.

• practice — not a real PYQ

  At the highest point of the path of a ball thrown vertically upward

  • (a)both velocity and acceleration are zero
  • (b)velocity is zero but acceleration is 9·8 m/s² downward
  • (c)acceleration is zero but velocity is maximum
  • (d)both velocity and acceleration are maximum

  Answer(b) velocity is zero but acceleration is 9·8 m/s² downward — gravity never switches off.