Two balls, A and B, are thrown simultaneously, A vertically upward with a speed of 20 m/s from the ground and B vertically downward from a height of 40 m with the same speed and along the same line of motion. At what points do the two balls collide by taking acceleration due to gravity as 9·8 m/s² ?
- (a)The balls will collide after 3s at a height of 30·2 m from the ground
- (b)The balls will collide after 2s at a height of 20·1 m from the ground
- (c)The balls will collide after 1s at a height of 15·1 m from the ground
- (d)The balls will collide after 5s at a height of 20 m from the ground
Correct — C, after 1 second at a height of 15·1 m. Measure heights upward from the ground. Ball A rises from zero, so its height at time t is 20t − 4·9t². Ball B falls from 40 m, so its height is 40 − 20t − 4·9t². They meet when the two heights are equal: 20t − 4·9t² = 40 − 20t − 4·9t². The 4·9t² term cancels on both sides, leaving 40t = 40, so t = 1 s. Substituting back, ball A is at 20(1) − 4·9(1)² = 15·1 m, and ball B is at 40 − 20 − 4·9 = 15·1 m as well, which confirms the meeting point. The shortcut is the cancellation itself: both balls fall away from their straight-line paths by the same 4·9t², so relative to each other they simply close a 40 m gap at a combined 40 m/s, which takes exactly one second.
- (a)The balls will collide after 3s at a height of 30·2 m from the ground — By 3 seconds ball B would have to be at 40 − 60 − 44·1 metres, which is far below the ground, so it has long since landed. Ball A at 3 seconds is at 15·9 m, nowhere near the quoted 30·2 m.
- (b)The balls will collide after 2s at a height of 20·1 m from the ground — The most tempting wrong option, because at 2 seconds ball A really is near this height, at 20·4 m. But ball B by then is already below ground level, so no collision is possible. The mid-air height of one ball is not enough — both have to be at the same place at the same instant.
- (d)The balls will collide after 5s at a height of 20 m from the ground — Ball A returns to the ground after 2 × 20 ÷ 9·8, which is about 4·1 seconds, so at 5 seconds neither ball is still in flight. The neat figure of 20 m is bait for anyone who halves the 40 m drop without doing any kinematics.
Both balls are in free flight, so both carry the same downward acceleration g regardless of their speed or direction of travel. Writing the position of each with the standard relation s = ut + ½at², and then setting the two positions equal, gives the time of meeting. Because the acceleration term is identical for both bodies, it cancels out of the equation — which is the whole reason this problem collapses to a single line of arithmetic.
The cancellation is worth understanding rather than memorising. In the frame of one ball, the other has zero relative acceleration and a constant relative velocity equal to the sum of the two speeds, here 20 + 20 = 40 m/s. Two bodies closing a 40 m gap at a constant 40 m/s meet after one second, and gravity affects only where in space that meeting happens, not when. Note also that the question quotes g as 9·8 m/s², so the decimal 15·1 falls out exactly; with g taken as 10, the height would come to 15 m instead.
- For a body in free flight, height above the launch point is ut − ½gt² when thrown up and −ut − ½gt² relative to the release point when thrown down.
- Two bodies in free flight have zero relative acceleration, so their relative motion is at constant velocity.
- Here the relative approach speed is 20 + 20 = 40 m/s over a gap of 40 m, giving a meeting time of exactly 1 second.
- Ball A thrown up at 20 m/s reaches its highest point after about 2·04 s and returns to the ground after about 4·08 s, so both balls are still in flight at 1 second — which the wrong options are not.
- Checking only one ball's height and ignoring whether the other is still in the air.
- Forgetting to reverse the sign of the initial velocity for the ball thrown downward.
- Using g = 10 when the question explicitly supplies 9·8, which shifts the answer from 15·1 m to 15 m.
NDA sets one or two worked kinematics numericals every paper, usually solvable in under a minute if you write both position equations and look for a term that cancels.
A ball is thrown vertically upward with a speed of 40 m/s. The time taken by the ball to reach the maximum height would be approximately
- (a) 2 s
- (b) 3 s
- (c) 4 s
- (d) 5 s
Answer(c) 4 s
The single-ball version of the same physics. Getting the time of ascent right is what tells you whether ball A in this 2016 item is even still airborne at the times the wrong options propose.
A rigid body of mass 2 kg is dropped from a stationary balloon kept at a height of 50 m from the ground. The speed of the body when it just touches the ground and the total energy when it is dropped from the balloon are respectively (acceleration due to gravity = 9·8 m/s²)
- (a) 980 m s⁻¹ and 980 J
- (b) √980 m s⁻¹ and √980 J
- (c) 980 m s⁻¹ and √980 J
- (d) √980 m s⁻¹ and 980 J
Answer(d) √980 m s⁻¹ and 980 J
A free-fall numerical with the same g = 9·8 convention, solved through energy instead of kinematics — a good reminder that these items often have a second, shorter route.
- practice — not a real PYQ
A ball is dropped from a height of 45 m at the same instant that another is thrown vertically upward from the ground at 30 m/s along the same line. Taking g = 10 m/s², after how long do they meet?
- (a)1·0 s
- (b)1·5 s
- (c)2·0 s
- (d)3·0 s
Answer(b) 1·5 s — the relative approach speed is 30 m/s over a 45 m gap, since the gravity terms cancel between the two bodies.
- practice — not a real PYQ
A ball thrown vertically upward from the ground with speed u returns to the ground after a total time of
- (a)u/g
- (b)u/2g
- (c)2u/g
- (d)u²/2g
Answer(c) 2u/g — it takes u/g to rise to the top and the same time again to fall back, the height reached being u²/2g.